OUTPUT of Python run 5 of at most 300:
```
Activation coverage:
1 nonzero neurons 3584 RMS first128/rest 0.06130542978644371 0.009526209905743599
11 nonzero neurons 3584 RMS first128/rest 0.052453044801950455 0.01794137805700302
18 nonzero neurons 3584 RMS first128/rest 0.18930694460868835 0.03775868937373161
23 nonzero neurons 3584 RMS first128/rest 0.7207475900650024 0.1179308220744133
Correct example IDs/tokens: [('gsm8k_train_2812', 1771), ('gsm8k_train_5235', 4157), ('gsm8k_train_3565', 2413), ('gsm8k_train_1949', 5441), ('gsm8k_train_966', 15404), ('gsm8k_train_3223', 8599), ('gsm8k_train_2226', 842), ('gsm8k_train_2527', 1262), ('gsm8k_train_1372', 6882), ('gsm8k_train_521', 2562), ('gsm8k_train_979', 10079), ('gsm8k_train_1690', 7231)]

PROBLEM gsm8k_train_2478 In a game, Samanta has 8 more points than Mark, and Mark has 50% more points than Eric. Eric has 6 points. How many points do Samanta, Mark, and Eric have in total?
offset 0 Here's a thinking process that leads to the solution:

1.  **Analyze the Request:**
    *   **Characters:** Samanta, Mark, Eric.
    *   **Relationships:**
        *   Samanta has 8 more points than Mark ($Samanta = Mark + 8$).
        *   Mark has 50% more points than Eric ($Mar
offset 350 $).
    *   **Goal:** Calculate the total number of points for Samanta, Mark, and Eric combined.

2.  **Step-by-Step Breakdown & Calculation:**

    *   *Given:* $Eric = 6$.

    *   *Find Mark's Points:*
        *   Rule: "Mark has 50% more points than Eric."
        *   Math: $
offset 700 5(6)$.
        *   Calculate: $Mark = 6 + 3 = 9$.
        *   *Self-Correction/Verification:* Wait, does "50% more" mean $1.5 \times$ or $0.5 \times$? Usually, "X is Y% more than Z" means $Z + (Y\% \text{ of } Z)$.
            *   Interpretation A: $Mark = 6 + 0.5 \times 6 = 9$.

offset 1200          *   Calculation: $6 + 0.5(6) = 6 + 3 = 9$. So Mark = 9.

    *   *Find Samanta's Points:*
        *   Rule: "Samanta has 8 more points than Mark."
        *   Math: $Samanta = Mark + 8$.
        *   Substitute: $Samanta = 9 + 8$.
        *   Calculation: $Samanta = 17$.


PROBLEM gsm8k_train_3192 Clara is climbing to the top of a historical tower with stone stairs. Each level has eight huge steps in the stairs. Every step is made up of three massive blocks of stone. By the time she reaches the top, she has climbed past 96 blocks of stone. How many levels are there in the tower?
offset 0 Here's a thinking process that leads to the solution:

1.  **Analyze the Problem:**
    *   Goal: Find the number of levels (steps) in the tower.
    *   Condition 1: There are huge steps.
    *   Condition 2: Each level has 8 huge steps.
    *   Condition 3: Each step is made up
offset 350  of stone.

2.  **Identify Variables and Relationships:**
    *   Let $n$ be the number of levels.
    *   The problem statement says "Each level has eight huge steps". This implies there are $n$ levels, where each level consists of some number of steps. However, usually, in thes
offset 700 locks per level.
    *   Let's look at the wording carefully: "Every step is made up of three massive blocks of stone."
    *   So, the structure is likely hierarchical.
        *   Level 1: 1 step? Or does it mean one level = 8 steps?
        *   Let's assume standard interpreta
offset 1200 By the time she reaches the top... she has climbed past 96 blocks of stone."
    *   Wait, let's re-read the constraint on "blocks of stone".
    *   Usually, "climbing a tower with stone stairs" implies climbing from bottom to top.
    *   If she climbs *one* level, she traverse

PROBLEM gsm8k_train_5272 Sweet Hope hosts a casino with 3 roulette tables labeled A, B, and C. At 6:00 pm, Table B had twice as much money on it as table C, which had $20 more than table A. If there was $40 on table A, how much money was there on all the tables?
offset 0 Here's a thinking process that leads to the solution:

1.  **Analyze the Problem Statement:**
    *   **Entities:** A casino with 3 roulette tables (A, B, C).
    *   **Time Context:** 6:00 pm.
    *   **Relationships given:**
        *   Table B had twice as much money on it as 
offset 350 e A ($C = A + 20$).
        *   There was $40 on table A ($A = 40$).
    *   **Goal:** Find the total amount of money on all three tables ($A + B + C$).

2.  **Break Down the Variables:**
    *   Let $x$ be the amount of money on Table A.
    *   Let $y$ be the amount of money on
offset 700 tract Equations from the Text:**
    *   "There was $40 on table A":
        $$x = 40$$
    *   "Table C had $20 more than table A":
        $$z = x + 20$$
    *   "Table B had twice as much money on it as table C":
        $$y = 2z$$

4.  **Perform Calculations:**
    *   From s
offset 1200 rst relationship given in the text: "Table B had twice as much money on it as table C". Wait, let me re-read carefully.
        *   "Table B had twice as much money on it as table C" -> $B = 2 \times C$.
        *   "Table C had $20 more than table A" -> $C = A + 20$.
        *  

PROBLEM math_train_precalculus_647 Let $\mathbf{a} = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix}.$  Find the vector $\mathbf{b}$ such that $\mathbf{a} \cdot \mathbf{b} = 11$ and
\[\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -13 \\ -9 \\ 7 \end{pmatrix}.\]
offset 0 The user wants to find a vector $\mathbf{b}$ given two conditions:
1. The dot product $\mathbf{a} \cdot \mathbf{b} = 11$.
2. The cross product $\mathbf{a} \times \mathbf{b} = \mathbf{c}$, where $\mathbf{c} = \begin{pmatrix} -13 \\ -9 \\ 7 \end{pmatrix}$.

Let the vectors be in $\
offset 350  (b_1, b_2, b_3)$
$\mathbf{c} = (-13, -9, 7)$

We have the equations:
1. $\mathbf{a} \cdot \mathbf{b} = 2b_1 + 1b_2 + 5b_3 = 11$
2. $\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 5 \\ b_1 & b_2 & b_3 \end{vmatrix} = -13\mathbf{i} -
offset 700  product.
$x_{cross} = a_2 b_3 - a_3 b_2 = 1 \cdot b_3 - 5 \cdot b_2 = b_3 - 5b_2$
$y_{cross} = a_3 b_1 - a_1 b_3 = 5 \cdot b_1 - 2 \cdot b_3 = 5b_1 - 2b_3$
$z_{cross} = a_1 b_2 - a_2 b_1 = 2 \cdot b_2 - 1 \cdot b_1 = 2b_2 - b_1$

So we have the system:
$b_3 - 5b_2 = -13$  (Eq 1'
offset 1200  $b_1 = 2b_2 - 7$

Now substitute $b_1$ and $b_3$ from these expressions into the third equation.
Wait, I used $b_1, b_2, b_3$ as variables. Let's solve for $b_1, b_2, b_3$ directly or use elimination.
We have:
1) $b_1 = 2b_2 - 7$
2) $b_3 = 5b_2 - 13$
3) $b_1 = 2b_2 - 7$ (from pr

Linear-attention decay parameters (head order):
0 A_log [0.25999999046325684, -2.0299999713897705, -2.119999885559082, -0.5699999928474426, -2.609999895095825, -2.25, -1.7999999523162842, -2.9800000190734863, -6.590000152587891, -0.6600000262260437, -1.7799999713897705, -0.9100000262260437, -1.649999976158142, -1.4500000476837158, -1.899999976158142, -2.609999895095825] dt_bias [2.549999952316284, -5.590000152587891, -4.619999885559082, 8.380000114440918, 3.640000104904175, -5.03000020980835, -5.059999942779541, -1.0800000429153442, -13.0600004196167, -5.440000057220459, -4.809999942779541, 6.440000057220459, -4.940000057220459, 7.590000152587891, -6.059999942779541, -3.2300000190734863]
4 A_log [-0.8799999952316284, -0.029999999329447746, -1.159999966621399, -0.6399999856948853, -0.8999999761581421, -2.5899999141693115, -1.350000023841858, -0.1599999964237213, 0.6000000238418579, -1.309999942779541, -0.9100000262260437, -0.6200000047683716, -2.9800000190734863, 0.5799999833106995, -1.3899999856948853, -2.7799999713897705] dt_bias [0.03999999910593033, 2.25, -4.46999979019165, 1.8899999856948853, -5.380000114440918, -5.53000020980835, -5.28000020980835, 0.9300000071525574, 0.10999999940395355, -3.049999952316284, -0.25999999046325684, -5.5, -3.3399999141693115, -5.25, -0.11999999731779099, -4.03000020980835]
8 A_log [-1.0499999523162842, -1.2599999904632568, -1.2100000381469727, -1.8899999856948853, -1.7999999523162842, -0.5, -1.4900000095367432, -0.9800000190734863, -0.14000000059604645, -2.299999952316284, -1.399999976158142, -1.0199999809265137, -0.9300000071525574, -2.7200000286102295, -0.41999998688697815, -1.0199999809265137] dt_bias [-0.8799999952316284, -1.9800000190734863, -4.46999979019165, -3.7300000190734863, -3.0199999809265137, 0.20999999344348907, -4.21999979019165, -0.9100000262260437, -2.5, -4.5, -0.15000000596046448, 0.25, -4.690000057220459, -4.059999942779541, -1.1200000047683716, -0.20000000298023224]
12 A_log [-1.600000023841858, -0.5299999713897705, -2.2699999809265137, -1.2599999904632568, -1.3200000524520874, 0.0, -2.0299999713897705, -3.0299999713897705, -2.200000047683716, -0.7699999809265137, -1.7999999523162842, -0.05000000074505806, -1.7799999713897705, -2.0799999237060547, -2.049999952316284, -1.1100000143051147] dt_bias [-5.0, -1.1699999570846558, -3.8399999141693115, -4.75, -1.9800000190734863, -6.659999847412109, -3.859999895095825, -3.4700000286102295, -4.590000152587891, -1.9199999570846558, -1.340000033378601, -0.5699999928474426, -4.96999979019165, -5.059999942779541, -1.2899999618530273, -1.2000000476837158]
16 A_log [-1.590000033378601, -0.8999999761581421, -2.119999885559082, -1.5299999713897705, -2.0199999809265137, -0.27000001072883606, -0.6299999952316284, -2.0199999809265137, -0.8999999761581421, -2.1700000762939453, -1.4600000381469727, -1.399999976158142, -1.5700000524520874, -1.7999999523162842, -1.75, -0.6700000166893005] dt_bias [-4.03000020980835, -0.6800000071525574, -4.25, -0.17000000178813934, -4.71999979019165, -0.10999999940395355, 1.4800000190734863, -3.7799999713897705, 0.8700000047683716, -4.090000152587891, 2.1600000858306885, 0.8999999761581421, -5.25, -4.309999942779541, -4.309999942779541, 0.3700000047683716]
20 A_log [-1.5, -1.75, -1.7000000476837158, -1.25, -1.4600000381469727, -0.6600000262260437, -0.8199999928474426, -1.2200000286102295, -0.9900000095367432, -0.7099999785423279, -1.3600000143051147, -0.7900000214576721, -1.6799999475479126, -1.75, -1.2799999713897705, -1.7699999809265137] dt_bias [-3.190000057220459, -0.3700000047683716, -4.659999847412109, 1.909
[output cut at 10,000 characters]
```