#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < int(n); ++i)

using matrix = vector<vector<int>>;

void matmul_assign(matrix& A, const matrix& B, int mod) {
    int N = int(A.size());
    matrix C(N, vector<int>(N));
    rep(k, N) rep(i, N) rep(j, N) {
        if ((C[i][j] += A[i][k] * B[k][j] % mod) >= mod) C[i][j] -= mod;
    }
    A.swap(C);
}

void matpow_assign(matrix& A, int e, int mod) {
    int N = int(A.size());
    matrix B(N, vector<int>(N));
    rep(i, N) B[i][i] = 1;
    for (; e > 0; e >>= 1, matmul_assign(A, A, mod)) if (e & 1) matmul_assign(B, A, mod);
    A.swap(B);
}

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

    for (int N, M, A, B, C, T;
         cin >> N >> M >> A >> B >> C >> T, N | M | A | B | C | T;) {
        vector<int> S(N);
        for (int& e : S) cin >> e;
        matrix R(N, vector<int>(N));
        rep(i, N) {
            if (i > 0) R[i][i - 1] = A;
            R[i][i] = B;
            if (i < N - 1) R[i][i + 1] = C;
        }
        matpow_assign(R, T, M);
        rep(i, N) {
            if (i) cout << ' ';
            int s = 0;
            rep(j, N) if ((s += R[i][j] * S[j] % M) >= M) s -= M;
            cout << s;
        }
        cout << endl;
    }
}
### Prompt

Create a solution in Cpp for the following problem:

There is a one-dimensional cellular automaton consisting of N cells. Cells are numbered from 0 to N − 1.

Each cell has a state represented as a non-negative integer less than M. The states of cells evolve through discrete time steps. We denote the state of the i-th cell at time t as S(i, t). The state at time t + 1 is defined by the equation

S(i, t + 1) = (A × S(i − 1, t) + B × S(i, t) + C × S(i + 1, t)) mod M,         (1)

where A, B and C are non-negative integer constants. For i < 0 or N ≤ i, we define S(i, t) = 0.

Given an automaton definition and initial states of cells, your mission is to write a program that computes the states of the cells at a specified time T.



Input

The input is a sequence of datasets. Each dataset is formatted as follows.

N M A B C T
S(0, 0) S(1, 0) ... S(N − 1, 0)


The first line of a dataset consists of six integers, namely N, M, A, B, C and T. N is the number of cells. M is the modulus in the equation (1). A, B and C are coefficients in the equation (1). Finally, T is the time for which you should compute the states.

You may assume that 0 < N ≤ 50, 0 < M ≤ 1000, 0 ≤ A, B, C < M and 0 ≤ T ≤ 109.

The second line consists of N integers, each of which is non-negative and less than M. They represent the states of the cells at time zero.

A line containing six zeros indicates the end of the input.

Output

For each dataset, output a line that contains the states of the cells at time T. The format of the output is as follows.

S(0, T) S(1, T) ... S(N − 1, T)

Each state must be represented as an integer and the integers must be separated by a space.

Example

Input

5 4 1 3 2 0
0 1 2 0 1
5 7 1 3 2 1
0 1 2 0 1
5 13 1 3 2 11
0 1 2 0 1
5 5 2 0 1 100
0 1 2 0 1
6 6 0 2 3 1000
0 1 2 0 1 4
20 1000 0 2 3 1000000000
0 1 2 0 1 0 1 2 0 1 0 1 2 0 1 0 1 2 0 1
30 2 1 0 1 1000000000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
30 2 1 1 1 1000000000
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
30 5 2 3 1 1000000000
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0


Output

0 1 2 0 1
2 0 0 4 3
2 12 10 9 11
3 0 4 2 1
0 4 2 0 4 4
0 376 752 0 376 0 376 752 0 376 0 376 752 0 376 0 376 752 0 376
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 3 2 2 2 3 3 1 4 3 1 2 3 0 4 3 3 0 4 2 2 2 2 1 1 2 1 3 0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < int(n); ++i)

using matrix = vector<vector<int>>;

void matmul_assign(matrix& A, const matrix& B, int mod) {
    int N = int(A.size());
    matrix C(N, vector<int>(N));
    rep(k, N) rep(i, N) rep(j, N) {
        if ((C[i][j] += A[i][k] * B[k][j] % mod) >= mod) C[i][j] -= mod;
    }
    A.swap(C);
}

void matpow_assign(matrix& A, int e, int mod) {
    int N = int(A.size());
    matrix B(N, vector<int>(N));
    rep(i, N) B[i][i] = 1;
    for (; e > 0; e >>= 1, matmul_assign(A, A, mod)) if (e & 1) matmul_assign(B, A, mod);
    A.swap(B);
}

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

    for (int N, M, A, B, C, T;
         cin >> N >> M >> A >> B >> C >> T, N | M | A | B | C | T;) {
        vector<int> S(N);
        for (int& e : S) cin >> e;
        matrix R(N, vector<int>(N));
        rep(i, N) {
            if (i > 0) R[i][i - 1] = A;
            R[i][i] = B;
            if (i < N - 1) R[i][i + 1] = C;
        }
        matpow_assign(R, T, M);
        rep(i, N) {
            if (i) cout << ' ';
            int s = 0;
            rep(j, N) if ((s += R[i][j] * S[j] % M) >= M) s -= M;
            cout << s;
        }
        cout << endl;
    }
}
```