<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

386. Happy Birthday, Jedi Knight!
Time limit per test: 0.25 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

Jedi Knight has sneaked into the new model of Death Star. He is searching for very important enemy documents. For each document he finds he will get a certain amount of money from the Jedi Council. A new model of Death Star has the form of an n-dimensional parallelepiped. Its edges are parallel to vectors v1,..., vn with integer coordinates and have the lengths equal to lengths of the corresponding vectors. There is one document located in every integer point inside or on the border of the Death Star. If for any k (0 ≤ k ≤ n) a document is inside some k-dimensional facet of the parallelepiped, but either k = 0 or it is not inside any (k-1)-dimensional facet, then its price is 2k. Your task is to calculate the total amount of money Jedi can earn if he gets all the documents on the Death Star. The answer can be enormous, but Jedi isn't afraid of this fact, so you should output it modulo prime number p.
Input
The first line of the input file contains numbers n and p (2 ≤ n ≤ 50, ). The next n lines contain description of vectors vi. Each of these lines contains n integer numbers aij (0 ≤ aij ≤ p-1) — coordinates of the vector vi. It is guaranteed that these vectors are linearly independent.
Output
Output must contain one number — the answer modulo p.
Example(s)
sample input
sample output
2 43
1 0
0 1
4

<|response|>
## 1. Abridged problem statement

Given `n` linearly independent integer vectors `v1, v2, ..., vn` in `n`-dimensional space, they define a parallelepiped

\[
P=\left\{\sum_{i=1}^{n} t_i v_i \mid 0 \le t_i \le 1\right\}.
\]

There is one document at every integer lattice point inside or on the border of `P`.

For a point

\[
x=\sum_{i=1}^{n} t_i v_i,
\]

let `k` be the number of coordinates `t_i` such that

\[
0 < t_i < 1.
\]

Then the document price is `2^k`.

Compute the total price of all documents modulo a prime number `p`.

Constraints include:

- `2 <= n <= 50`
- vectors are linearly independent
- coordinates are integers modulo `p`

---

## 2. Key observations needed to solve the problem

### Observation 1: Facet dimension equals number of non-boundary parameters

Since the vectors are linearly independent, every point in the parallelepiped has a unique representation:

\[
x=\sum_{i=1}^{n} t_i v_i,\qquad 0 \le t_i \le 1.
\]

If `t_i = 0` or `t_i = 1`, the point lies on a boundary wall in direction `i`.

If `0 < t_i < 1`, the point is free in that direction.

Therefore, the smallest facet containing the point has dimension

\[
k = \#\{i : 0 < t_i < 1\}.
\]

So the price is exactly

\[
2^k.
\]

---

### Observation 2: Use the half-open parallelepiped

Define the half-open parallelepiped:

\[
H=\left\{\sum_{i=1}^{n} t_i v_i \mid 0 \le t_i < 1\right\}.
\]

The integer vectors `v_i` generate a lattice:

\[
L=\left\{\sum_{i=1}^{n} z_i v_i \mid z_i \in \mathbb Z\right\}.
\]

The half-open parallelepiped `H` is a fundamental domain of this lattice.

The number of integer points in `H` is equal to the index of this lattice in `\mathbb Z^n`, which is

\[
|H \cap \mathbb Z^n| = |\det V|,
\]

where `V` is the matrix formed by the vectors.

---

### Observation 3: Every half-open representative contributes the same amount

Take an integer point in `H`:

\[
y=\sum_{i=1}^{n} s_i v_i,\qquad 0 \le s_i < 1.
\]

Suppose exactly `z` of the parameters `s_i` are zero.

For every zero coordinate, in the closed parallelepiped we may choose either:

- `t_i = 0`
- `t_i = 1`

So this representative corresponds to

\[
2^z
\]

points in the closed parallelepiped.

For all these points:

- the `z` coordinates are on the boundary,
- the remaining `n - z` coordinates are strictly inside.

Thus each such point has price

\[
2^{n-z}.
\]

Therefore the whole group contributes

\[
2^z \cdot 2^{n-z} = 2^n.
\]

So every integer point in `H` contributes exactly `2^n` to the total answer.

Since there are `|\det V|` such points, the answer is

\[
\boxed{2^n \cdot |\det V|}
\]

modulo `p`.

---

## 3. Full solution approach based on the observations

We only need to compute:

\[
\text{answer} = 2^n \cdot |\det V| \pmod p.
\]

The determinant may be enormous, so we compute it modulo `p`.

However, modulo arithmetic gives us `det(V) mod p`, while we need `|det(V)| mod p`.

So:

1. Compute `det(V) mod p` using Gaussian elimination modulo `p`.
2. Determine the sign of the real determinant.
3. If the determinant is negative, replace `det` by `-det mod p`.
4. Multiply by `2^n mod p`.

If

\[
\det V \equiv 0 \pmod p,
\]

then the answer is zero regardless of the determinant sign.

### Complexity

Gaussian elimination takes:

\[
O(n^3)
\]

which is easily fine for `n <= 50`.

---

## 4. C++ implementation with detailed comments

```cpp
#include <bits/stdc++.h>
using namespace std;

using int64 = long long;

int n;
int64 p;
vector<vector<int64>> a;

/*
    Multiplies two numbers modulo p.

    We use __int128 because p can be large enough that
    x * y may overflow 64-bit integer.
*/
int64 mul_mod(int64 x, int64 y) {
    return (int64)((__int128)x * y % p);
}

/*
    Fast modular exponentiation.

    Computes base^exp modulo p in O(log exp).
*/
int64 pow_mod(int64 base, int64 exp) {
    int64 result = 1 % p;
    base %= p;

    while (exp > 0) {
        if (exp & 1) {
            result = mul_mod(result, base);
        }

        base = mul_mod(base, base);
        exp >>= 1;
    }

    return result;
}

/*
    Computes det(a) modulo p using Gaussian elimination over the field mod p.

    Since p is prime, every nonzero element modulo p has an inverse.
*/
int64 determinant_mod_p(vector<vector<int64>> mat) {
    int64 det = 1 % p;

    for (int col = 0; col < n; col++) {
        int pivot = -1;

        /*
            Find a row with a nonzero value in the current column.
        */
        for (int row = col; row < n; row++) {
            if (mat[row][col] % p != 0) {
                pivot = row;
                break;
            }
        }

        /*
            If no pivot exists, determinant is zero modulo p.
        */
        if (pivot == -1) {
            return 0;
        }

        /*
            Swap pivot row into place.

            Swapping two rows changes determinant sign.
        */
        if (pivot != col) {
            swap(mat[pivot], mat[col]);
            det = (p - det) % p;
        }

        int64 pivot_value = mat[col][col] % p;

        /*
            Multiply determinant by current pivot.
        */
        det = mul_mod(det, pivot_value);

        /*
            Compute inverse of pivot modulo p.

            Fermat's little theorem:
                x^(p - 2) == x^(-1) mod p
            because p is prime.
        */
        int64 inv_pivot = pow_mod(pivot_value, p - 2);

        /*
            Eliminate entries below the pivot.
        */
        for (int row = col + 1; row < n; row++) {
            int64 factor = mul_mod(mat[row][col] % p, inv_pivot);

            for (int c = col; c < n; c++) {
                mat[row][c] =
                    (mat[row][c] - mul_mod(factor, mat[col][c]) + p) % p;
            }
        }
    }

    return det;
}

/*
    Determines the sign of the real determinant.

    We use long double Gaussian elimination with partial pivoting.

    The determinant sign is affected by:
    - row swaps,
    - signs of pivot elements.

    Row additions do not change determinant.
*/
int determinant_sign() {
    vector<vector<long double>> b(n, vector<long double>(n));

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            b[i][j] = (long double)a[i][j];
        }
    }

    int sign = 1;

    for (int col = 0; col < n; col++) {
        int pivot = col;

        /*
            Choose the row with the largest absolute value in this column.
            This improves numerical stability.
        */
        for (int row = col + 1; row < n; row++) {
            if (fabsl(b[row][col]) > fabsl(b[pivot][col])) {
                pivot = row;
            }
        }

        /*
            The vectors are linearly independent, so the determinant is nonzero.
        */
        if (fabsl(b[pivot][col]) == 0) {
            return 0;
        }

        if (pivot != col) {
            swap(b[pivot], b[col]);
            sign = -sign;
        }

        /*
            The pivot contributes its sign to the determinant.
        */
        if (b[col][col] < 0) {
            sign = -sign;
        }

        /*
            Eliminate below.
        */
        for (int row = col + 1; row < n; row++) {
            long double factor = b[row][col] / b[col][col];

            for (int c = col; c < n; c++) {
                b[row][c] -= factor * b[col][c];
            }
        }
    }

    return sign;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n >> p;

    a.assign(n, vector<int64>(n));

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> a[i][j];
        }
    }

    /*
        Mathematical result:

            answer = 2^n * |det(V)| mod p
    */

    int64 det_mod = determinant_mod_p(a);

    /*
        If det is divisible by p, then |det| mod p is also zero.
        In that case, sign does not matter.
    */
    if (det_mod != 0) {
        int sign = determinant_sign();

        /*
            If det(V) is negative, then:

                |det(V)| = -det(V)

            modulo p this means replacing det_mod by p - det_mod.
        */
        if (sign < 0) {
            det_mod = (p - det_mod) % p;
        }
    }

    int64 power_of_two = pow_mod(2, n);

    int64 answer = mul_mod(det_mod, power_of_two);

    cout << answer << '\n';

    return 0;
}
```

---

## 5. Python implementation with detailed comments

In Python, we can avoid floating-point sign detection by computing the determinant exactly.

A good exact determinant algorithm is the Bareiss algorithm, which is a fraction-free version of Gaussian elimination. It works entirely with integers and has `O(n^3)` complexity.

```python
import sys


def bareiss_det(matrix):
    """
    Computes the exact integer determinant of a square matrix.

    Bareiss algorithm is a fraction-free Gaussian elimination method.

    It keeps all intermediate values as integers and performs exact divisions.
    For n <= 50, this is practical in Python.
    """

    # Make a copy because the algorithm modifies the matrix.
    a = [row[:] for row in matrix]

    n = len(a)

    # Row swaps change determinant sign.
    sign = 1

    # Previous pivot used in the Bareiss formula.
    prev_pivot = 1

    for k in range(n - 1):
        pivot_row = k

        # Find a nonzero pivot.
        while pivot_row < n and a[pivot_row][k] == 0:
            pivot_row += 1

        # If no pivot exists, determinant is zero.
        if pivot_row == n:
            return 0

        # Swap pivot row into place.
        if pivot_row != k:
            a[k], a[pivot_row] = a[pivot_row], a[k]
            sign *= -1

        pivot = a[k][k]

        """
        Bareiss update:

            a[i][j] = (a[i][j] * pivot - a[i][k] * a[k][j]) / prev_pivot

        The division is exact for integer matrices.
        """
        for i in range(k + 1, n):
            for j in range(k + 1, n):
                a[i][j] = (
                    a[i][j] * pivot - a[i][k] * a[k][j]
                ) // prev_pivot

        # Clear entries below the pivot.
        for i in range(k + 1, n):
            a[i][k] = 0

        prev_pivot = pivot

    # For a 1x1 matrix, determinant is its only value.
    if n == 1:
        return a[0][0]

    return sign * a[n - 1][n - 1]


def main():
    data = sys.stdin.buffer.read().split()

    n = int(data[0])
    p = int(data[1])

    values = list(map(int, data[2:]))

    matrix = []
    index = 0

    for _ in range(n):
        row = values[index:index + n]
        matrix.append(row)
        index += n

    """
    Mathematical result:

        answer = 2^n * |det(V)| mod p
    """

    det = bareiss_det(matrix)

    abs_det_mod = abs(det) % p

    power_of_two = pow(2, n, p)

    answer = abs_det_mod * power_of_two % p

    print(answer)


if __name__ == "__main__":
    main()
```

The core idea is that all geometric complexity collapses into one simple formula:

\[
\boxed{\text{answer} = 2^n \cdot |\det V| \pmod p}
\]

where `|det V|` is the number of integer lattice representatives in the half-open parallelepiped.