<|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;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
int64_t p;
vector<vector<int64_t>> a;

void read() {
    cin >> n >> p;
    a.assign(n, vector<int64_t>(n));
    cin >> a;
}

int64_t mul_mod(int64_t x, int64_t y) { return (int64_t)((__int128)x * y % p); }

int64_t pow_mod(int64_t base, int64_t e) {
    int64_t r = 1 % p;
    base %= p;
    while(e > 0) {
        if(e & 1) {
            r = mul_mod(r, base);
        }

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

    return r;
}

void solve() {
    // Every point of the body can be written x = sum t_i v_i for one choice
    // of coefficients t_1..t_n; since the vectors are independent those
    // coefficients (call them the parameters of x) are unique. The point is
    // in the body exactly when every parameter is in [0, 1]: all-zero is the
    // origin corner, all-one is the opposite corner, and a parameter equal to
    // 0 or 1 means x sits on one of the flat walls of that direction.
    //
    // A facet is one of those flat boundary pieces: a corner (dimension 0),
    // an edge (dimension 1), a 2D face, and so on up to the solid body itself
    // (dimension n). To name a facet, pin some parameters to a wall (0 or 1)
    // and let the rest range over [0, 1]; the points you get form a facet
    // whose dimension is the number of parameters left free. The smallest
    // facet through a given x pins exactly the parameters that already sit at
    // 0 or 1 and frees the rest, so its dimension is the count of strictly
    // interior parameters
    // 
    //     k = #{ i : 0 < t_i < 1 }.
    //     
    // A corner has k = 0, a point in the solid interior has k = n, and the
    // document price 2^k is precisely the 2^0 .. 2^n scale in the statement.
    // The task is to add up 2^k over every integer point (every document)
    // whose parameters all lie in [0, 1].
    //
    // Adding 2^k over the whole body head-on is hard, because the points and
    // their prices both vary. So we compare against a region whose integer
    // points we can count exactly: the half-open box
    // 
    //     H = { sum t_i v_i : 0 <= t_i < 1 },
    //
    // i.e. the body with its far walls (the t_i = 1 ones) shaved off. Take a
    // copy of H and shift it by an integer combination of v1..vn. These
    // shifted copies cover space with no gaps or overlaps -- they tile it,
    // the way unit squares tile the plane -- and this works precisely because
    // v1..vn are independent: being a basis they give every point of space a
    // unique parameter vector, and peeling off the integer parts of those
    // parameters (which is an integer combination of the v_i) slides the
    // point into exactly one copy of H. Unique copy means total coverage with
    // no overlap, so it really is a tiling.
    //
    // Now count the grid points in one copy. Each copy is an integer
    // translate of H, and translating by an integer vector is a bijection of
    // the grid Z^n onto itself, so every copy holds the same number m of grid
    // points. Count the grid points of a large box two ways: directly it is
    // about (volume of the box), since the grid has one point per unit
    // volume; and it is also (number of copies inside) * m, with about
    // (volume of the box) / |det V| copies because each copy has volume
    // |det V|. Equating the two forces m = |det V|, so H holds exactly
    // |det V| grid points. (This is the geometry-of-numbers fact that a
    // sublattice's fundamental domain contains [Z^n : L] = |det V| grid
    // points; its two-dimensional area form is Pick's theorem.) The plan is
    // to charge the entire price sum to these |det V| points of H.
    //
    // Claim: group the body's integer points by the point of H they reduce
    // to, and every group sums to exactly 2^n in price. There are |det V|
    // groups, so the whole price sum is 2^n * |det V| -- the answer. The
    // per-group total is always the same because prices factor coordinate by
    // coordinate, and the only freedom inside a group is which wall a
    // boundary coordinate sits on.
    //
    // Reduce a body point x to its representative y in H like this: while some
    // parameter t_i equals 1, subtract v_i, which lowers that parameter to 0,
    // keeps x in the body, and ends at a y with every parameter in [0, 1).
    // Conversely, the body points that reduce to a fixed y are exactly those
    // built from y by taking each coordinate where y is 0 and setting it to 0
    // or 1 (setting it to 1 adds v_i); a coordinate of y already in (0, 1)
    // cannot move, since raising it would pass 1. So if y has z zero
    // coordinates, its group is the 2^z independent 0/1 choices on them.
    //
    // Now total the group. Write the price 2^k as one factor per coordinate:
    // 2 if that coordinate is strictly interior, 1 if it sits on a wall (0 or
    // 1). Summed over the group each coordinate contributes a factor of 2 -- a
    // coordinate with y in (0, 1) is interior in every member (factor 2), and
    // a coordinate with y = 0 runs over both walls 0 and 1 (factors 1 + 1 =
    // 2). The group therefore totals 2^n for every y. Concretely, the group
    // has 2^z members each of price 2^(n - z), and 2^z * 2^(n - z) = 2^n.
    //
    // With |det V| groups each worth 2^n, the answer is 2^n * |det V| mod p.
    //
    // Gaussian elimination over F_p gives det V mod p but only up to sign,
    // and the entries are too large to take an exact big-integer det, so the
    // true sign is read off a plain floating-point LU with partial pivoting
    // and used to turn the residue into the nonnegative |det V|. When det V
    // is a multiple of p the residue is already 0 and the sign does not
    // matter.

    vector<vector<int64_t>> mat = a;
    int64_t det = 1 % p;
    for(int col = 0; col < n; col++) {
        int piv = -1;
        for(int r = col; r < n; r++) {
            if(mat[r][col] % p != 0) {
                piv = r;
                break;
            }
        }

        if(piv == -1) {
            det = 0;
            break;
        }

        if(piv != col) {
            swap(mat[piv], mat[col]);
            det = (p - det) % p;
        }

        det = mul_mod(det, mat[col][col] % p);
        int64_t inv = pow_mod(mat[col][col], p - 2);
        for(int r = col + 1; r < n; r++) {
            int64_t factor = mul_mod(mat[r][col] % p, inv);
            for(int c = col; c < n; c++) {
                mat[r][c] =
                    (mat[r][c] - mul_mod(factor, mat[col][c]) % p + p) % p;
            }
        }
    }

    if(det != 0) {
        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 piv = col;
            for(int r = col + 1; r < n; r++) {
                if(fabsl(b[r][col]) > fabsl(b[piv][col])) {
                    piv = r;
                }
            }

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

            if(b[col][col] < 0) {
                sign = -sign;
            }

            for(int r = col + 1; r < n; r++) {
                long double factor = b[r][col] / b[col][col];
                for(int c = col; c < n; c++) {
                    b[r][c] -= factor * b[col][c];
                }
            }
        }

        if(sign < 0) {
            det = (p - det) % p;
        }
    }

    cout << mul_mod(det, pow_mod(2, n)) << '\n';
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    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.
