## 1. Abridged problem statement

You are given `n` linearly independent integer vectors `v1, ..., vn` in `n`-dimensional space. They form the 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 boundary of `P`.

For a document point, let `k` be the dimension of the smallest facet of the parallelepiped containing it. Equivalently, if

\[
x=\sum t_i v_i,
\]

then `k` is the number of parameters `t_i` strictly between `0` and `1`. The document price is `2^k`.

Compute the sum of all document prices modulo a given prime `p`.

---

## 2. Detailed editorial

Let the parallelepiped be generated by the independent integer vectors

\[
v_1,\dots,v_n.
\]

Every point inside the parallelepiped has a unique representation

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

Because the vectors are linearly independent, the parameters `t_i` are unique.

For such a point:

- 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 interior in direction `i`.

Therefore the dimension `k` of the smallest facet containing the point is

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

The price of the document is then

\[
2^k.
\]

So the direct problem is:

\[
\sum_{x \in P \cap \mathbb Z^n} 2^{\#\{i:0<t_i<1\}}.
\]

---

### Key idea: half-open fundamental 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 sublattice

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

Because the `v_i` are integer vectors, `L` is a sublattice of `\mathbb Z^n`.

The half-open parallelepiped `H` is a fundamental domain of this lattice. That means every point in space can be uniquely shifted by an element of `L` into `H`.

In particular, the integer lattice points in `H` represent the cosets of

\[
\mathbb Z^n / L.
\]

The number of such cosets is the lattice index

\[
[\mathbb Z^n : L] = |\det V|,
\]

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

Thus:

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

---

### Grouping points of the closed parallelepiped

Now consider the original closed parallelepiped

\[
P = \left\{\sum t_i v_i \mid 0 \le t_i \le 1\right\}.
\]

Every integer point in `P` can be reduced to a unique integer point in `H` by replacing every coordinate `t_i = 1` with `t_i = 0`, i.e. subtracting `v_i`.

So we group points of `P` by their representative in `H`.

Take one representative

\[
y = \sum s_i v_i,\qquad 0 \le s_i < 1.
\]

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

For each coordinate where `s_i = 0`, we have two possibilities in the closed parallelepiped:

- keep `t_i = 0`;
- change it to `t_i = 1`.

For coordinates with `0 < s_i < 1`, we cannot change them, because adding `1` would exceed the allowed range.

Therefore this group contains exactly

\[
2^z
\]

points.

For every point in this group:

- the `z` changed/free zero-coordinates are always on a wall, either `0` or `1`;
- the remaining `n - z` coordinates are strictly inside.

So every point in the group has price

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

Hence the total contribution of this whole group is

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

This is the crucial observation: every representative in `H` contributes exactly `2^n`.

There are `|\det V|` representatives, so the total answer is

\[
2^n \cdot |\det V|.
\]

We only need this modulo `p`.

---

### Computing the answer

We need

\[
2^n \cdot |\det V| \pmod p.
\]

The determinant can be enormous, so the C++ solution computes:

1. `det(V) mod p` using Gaussian elimination modulo prime `p`;
2. the sign of the real determinant using floating-point Gaussian elimination;
3. converts `det mod p` into `|det| mod p`.

If

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

then the final answer is `0`, regardless of the sign.

Otherwise:

- if `det(V) > 0`, then `|det(V)| mod p = det mod p`;
- if `det(V) < 0`, then `|det(V)| mod p = -det mod p = p - det`.

Finally multiply by

\[
2^n \bmod p.
\]

Complexity:

\[
O(n^3)
\]

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

---

## 3. Provided C++ solution with detailed comments

```cpp
#include <bits/stdc++.h> // Includes almost all standard C++ headers.

using namespace std; // Allows using standard library names without std::.

// Output operator for pairs, useful for debugging/general convenience.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second; // Print first and second separated by space.
}

// Input operator for pairs.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second; // Read first and second.
}

// Input operator for vectors.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) { // Iterate by reference over all elements.
        in >> x;      // Read each element.
    }
    return in;        // Return input stream to allow chaining.
};

// Output operator for vectors.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {  // Iterate over all elements.
        out << x << ' '; // Print element followed by a space.
    }
    return out;       // Return output stream to allow chaining.
};

int n;                // Dimension of the parallelepiped.
int64_t p;            // Prime modulus.
vector<vector<int64_t>> a; // Matrix of vectors.

// Reads input.
void read() {
    cin >> n >> p; // Read dimension and modulus.

    a.assign(n, vector<int64_t>(n)); // Create an n by n matrix.

    cin >> a; // Read all matrix elements using overloaded vector input.
}

// Multiplies x and y modulo p safely.
// __int128 is used because x * y may overflow int64_t.
int64_t mul_mod(int64_t x, int64_t y) {
    return (int64_t)((__int128)x * y % p);
}

// Fast modular exponentiation: computes base^e mod p.
int64_t pow_mod(int64_t base, int64_t e) {
    int64_t r = 1 % p; // Result starts as multiplicative identity.

    base %= p; // Reduce base modulo p.

    while(e > 0) { // Binary exponentiation loop.
        if(e & 1) { // If current exponent bit is set...
            r = mul_mod(r, base); // Multiply result by current base.
        }

        base = mul_mod(base, base); // Square the base.
        e >>= 1;                    // Move to next exponent bit.
    }

    return r; // Return base^e mod p.
}

void solve() {
    /*
        Mathematical result:

        The total price equals

            2^n * |det(V)|.

        Reason:
        The half-open parallelepiped generated by the vectors contains exactly
        |det(V)| integer points. Each such point represents a group of points
        in the closed parallelepiped, and each group contributes exactly 2^n.
    */

    vector<vector<int64_t>> mat = a; // Copy matrix for modular Gaussian elimination.

    int64_t det = 1 % p; // Will store det(V) modulo p.

    // Gaussian elimination over the finite field modulo p.
    for(int col = 0; col < n; col++) {
        int piv = -1; // Pivot row index.

        // Search for a row with nonzero value in current column.
        for(int r = col; r < n; r++) {
            if(mat[r][col] % p != 0) { // Nonzero modulo p can be used as pivot.
                piv = r;               // Store pivot row.
                break;                 // Stop after first suitable pivot.
            }
        }

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

        // Move pivot row to current row if necessary.
        if(piv != col) {
            swap(mat[piv], mat[col]); // Swap rows.
            det = (p - det) % p;      // Row swap changes determinant sign.
        }

        // Multiply determinant by pivot element.
        det = mul_mod(det, mat[col][col] % p);

        // Modular inverse of pivot, using Fermat's little theorem.
        // Valid because p is prime and pivot is nonzero modulo p.
        int64_t inv = pow_mod(mat[col][col], p - 2);

        // Eliminate entries below pivot.
        for(int r = col + 1; r < n; r++) {
            // factor = mat[r][col] / mat[col][col] modulo p.
            int64_t factor = mul_mod(mat[r][col] % p, inv);

            // Subtract factor * pivot row from current row.
            for(int c = col; c < n; c++) {
                mat[r][c] =
                    (mat[r][c] - mul_mod(factor, mat[col][c]) % p + p) % p;
                    // Add p before modulo to avoid negative values.
            }
        }
    }

    /*
        At this point det is det(V) modulo p, but det(V) may be negative.
        We need |det(V)| modulo p.

        If det == 0 modulo p, then |det(V)| modulo p is also 0,
        so the sign does not matter.

        Otherwise, determine the sign of the real determinant using
        floating-point Gaussian elimination.
    */
    if(det != 0) {
        vector<vector<long double>> b(n, vector<long double>(n)); // Floating copy.

        // Copy integer matrix into long double matrix.
        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; // Tracks sign of determinant.

        // Gaussian elimination over real numbers with partial pivoting.
        for(int col = 0; col < n; col++) {
            int piv = col; // Start by assuming current row is pivot.

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

            // Swap pivot row into place if needed.
            if(piv != col) {
                swap(b[piv], b[col]); // Row swap.
                sign = -sign;         // Row swap flips determinant sign.
            }

            // The determinant is product of diagonal pivots after elimination.
            // If pivot is negative, it flips the sign.
            if(b[col][col] < 0) {
                sign = -sign;
            }

            // Eliminate rows below current pivot.
            for(int r = col + 1; r < n; r++) {
                long double factor = b[r][col] / b[col][col]; // Real elimination factor.

                for(int c = col; c < n; c++) {
                    b[r][c] -= factor * b[col][c]; // Subtract factor times pivot row.
                }
            }
        }

        // If the true determinant is negative, convert det modulo p to |det| modulo p.
        if(sign < 0) {
            det = (p - det) % p;
        }
    }

    // Final answer: |det(V)| * 2^n modulo p.
    cout << mul_mod(det, pow_mod(2, n)) << '\n';
}

int main() {
    ios_base::sync_with_stdio(false); // Fast C++ input/output.
    cin.tie(nullptr);                 // Do not flush cout before cin.

    int T = 1; // There is only one test case.

    // cin >> T; // Multiple test cases are not used in this problem.

    for(int test = 1; test <= T; test++) {
        read();  // Read input.
        solve(); // Solve and print answer.
    }

    return 0; // Successful program termination.
}
```

---

## 4. Python solution with detailed comments

This Python version uses the same formula:

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

Unlike the C++ version, Python can conveniently compute the determinant exactly using the Bareiss algorithm, a fraction-free Gaussian elimination method.

```python
import sys


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

    Bareiss algorithm is a fraction-free version of Gaussian elimination.
    It keeps all intermediate values as integers and has O(n^3) complexity.

    For n <= 50, this is easily practical in Python.
    """

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

    # Matrix size.
    n = len(a)

    # The determinant sign changes whenever we swap two rows.
    sign = 1

    # Previous pivot used for exact division.
    # In Bareiss algorithm, this starts as 1.
    prev = 1

    # Process columns 0 through n-2.
    # After that, the last diagonal value contains the determinant up to sign.
    for k in range(n - 1):
        # Find a nonzero pivot in column k at or below row k.
        pivot_row = k
        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 position k if needed.
        if pivot_row != k:
            a[k], a[pivot_row] = a[pivot_row], a[k]
            sign *= -1

        # Current pivot.
        pivot = a[k][k]

        # Update the remaining submatrix.
        for i in range(k + 1, n):
            for j in range(k + 1, n):
                # Bareiss update:
                #
                # a[i][j] = (a[i][j] * pivot - a[i][k] * a[k][j]) / prev
                #
                # The division is exact for integer matrices.
                a[i][j] = (a[i][j] * pivot - a[i][k] * a[k][j]) // prev

        # Entries below the pivot are no longer needed.
        for i in range(k + 1, n):
            a[i][k] = 0

        # Store this pivot for the next Bareiss division.
        prev = pivot

    # For a 1x1 matrix, determinant is just its only element.
    # The problem has n >= 2, but this keeps the function general.
    if n == 1:
        return a[0][0]

    # Last diagonal entry is determinant up to row-swap sign.
    return sign * a[n - 1][n - 1]


def main():
    # Read all input tokens.
    data = sys.stdin.buffer.read().split()

    # First token is n, second is p.
    n = int(data[0])
    p = int(data[1])

    # Read the n by n matrix.
    values = list(map(int, data[2:]))

    matrix = []
    index = 0

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

    # Compute exact determinant.
    det = bareiss_det(matrix)

    # We need |det| modulo p.
    abs_det_mod = abs(det) % p

    # Compute 2^n modulo p.
    pow2 = pow(2, n, p)

    # Final formula.
    answer = abs_det_mod * pow2 % p

    # Print answer.
    print(answer)


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

---

## 5. Compressed editorial

Every point of the parallelepiped has a unique representation

\[
x=\sum t_i v_i,\qquad 0\le t_i\le 1.
\]

The smallest facet dimension containing `x` is the number of coordinates with

\[
0<t_i<1.
\]

So the document price is `2^k`.

Consider the half-open parallelepiped

\[
H=\left\{\sum t_i v_i \mid 0\le t_i<1\right\}.
\]

It is a fundamental domain of the lattice generated by the integer vectors `v_i`. Therefore the number of integer points in `H` is

\[
|\det V|.
\]

Now group every integer point of the closed parallelepiped by reducing coordinates equal to `1` back to `0`, landing in `H`.

For a representative in `H`, suppose `z` of its coordinates are zero. Then it corresponds to `2^z` points in the closed parallelepiped, obtained by choosing each zero coordinate to be either `0` or `1`.

Each such point has exactly `n-z` strictly interior coordinates, hence price

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

So the whole group contributes

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

There are `|\det V|` groups, hence

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

Compute the determinant modulo `p`, adjust its sign to get the absolute determinant modulo `p`, and multiply by `2^n mod p`.