#include <cstdio>
#include <vector>
using namespace std;

typedef long long int ll;
ll MOD = 1 << 30;

// ?????????????????????
// Verified: TopCoder SRM 704 Div.2 (ModEquationEasy)
// ???????????? ??¨ ?´????(??°??????????????????)

// Matrix Library Begin (C++11)

template <typename T>
using Matrix = vector< vector<T> >;

template <typename T>
void init_mat(Matrix<T> &A, int h, int w) {
    A.resize(h, vector<T>(w, 0));
}

template <typename T>
Matrix<T> calc_mat(Matrix<T> A, Matrix<T> B) {
    Matrix<T> C(A.size(), vector<T>(B[0].size()));
    for(int i=0; i<A.size(); i++) {
        for(int k=0; k<B.size(); k++) {
            for(int j=0; j<B[0].size(); j++) {
                // C[i][j] += A[i][k] * B[k][j]; // mod??????
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD; // mod??????
            }
        }
    }
    return C;
}

template <typename T>
Matrix<T> mat_pow(Matrix<T> A, ll n) {
    Matrix<T> B(A.size(), vector<T>(A.size()));
    for(int i=0; i<A.size(); i++) B[i][i] = 1;
    while(n > 0) {
        if(n & 1) B = calc_mat(B, A);
        A = calc_mat(A, A);
        n >>= 1;
    }
    return B;
}

// Matrix Library End

int N, M, A, B, C, T;
int main() {
    while(1) {
        scanf("%d%d%d%d%d%d", &N, &M, &A, &B, &C, &T);
        if(N == 0) break;

        MOD = M;
        Matrix<ll> mat, vec;
        init_mat(mat, N, N);
        init_mat(vec, N, 1);

        for(int i=0; i<N; i++) {
            scanf("%lld", &vec[i][0]);
        }

        for(int i=0; i<N; i++) {
            for(int j=0; j<N; j++) {
                if(i-1 == j) mat[i][j] = A;
                if(i   == j) mat[i][j] = B;
                if(i+1 == j) mat[i][j] = C;
            }
        }

        Matrix<ll> ret = calc_mat(mat_pow(mat, T), vec);
        for(int i=0; i<N; i++) {
            if(i != 0) printf(" ");
            printf("%lld", ret[i][0]);
        }
        printf("\n");
    }
    return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 <cstdio>
#include <vector>
using namespace std;

typedef long long int ll;
ll MOD = 1 << 30;

// ?????????????????????
// Verified: TopCoder SRM 704 Div.2 (ModEquationEasy)
// ???????????? ??¨ ?´????(??°??????????????????)

// Matrix Library Begin (C++11)

template <typename T>
using Matrix = vector< vector<T> >;

template <typename T>
void init_mat(Matrix<T> &A, int h, int w) {
    A.resize(h, vector<T>(w, 0));
}

template <typename T>
Matrix<T> calc_mat(Matrix<T> A, Matrix<T> B) {
    Matrix<T> C(A.size(), vector<T>(B[0].size()));
    for(int i=0; i<A.size(); i++) {
        for(int k=0; k<B.size(); k++) {
            for(int j=0; j<B[0].size(); j++) {
                // C[i][j] += A[i][k] * B[k][j]; // mod??????
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD; // mod??????
            }
        }
    }
    return C;
}

template <typename T>
Matrix<T> mat_pow(Matrix<T> A, ll n) {
    Matrix<T> B(A.size(), vector<T>(A.size()));
    for(int i=0; i<A.size(); i++) B[i][i] = 1;
    while(n > 0) {
        if(n & 1) B = calc_mat(B, A);
        A = calc_mat(A, A);
        n >>= 1;
    }
    return B;
}

// Matrix Library End

int N, M, A, B, C, T;
int main() {
    while(1) {
        scanf("%d%d%d%d%d%d", &N, &M, &A, &B, &C, &T);
        if(N == 0) break;

        MOD = M;
        Matrix<ll> mat, vec;
        init_mat(mat, N, N);
        init_mat(vec, N, 1);

        for(int i=0; i<N; i++) {
            scanf("%lld", &vec[i][0]);
        }

        for(int i=0; i<N; i++) {
            for(int j=0; j<N; j++) {
                if(i-1 == j) mat[i][j] = A;
                if(i   == j) mat[i][j] = B;
                if(i+1 == j) mat[i][j] = C;
            }
        }

        Matrix<ll> ret = calc_mat(mat_pow(mat, T), vec);
        for(int i=0; i<N; i++) {
            if(i != 0) printf(" ");
            printf("%lld", ret[i][0]);
        }
        printf("\n");
    }
    return 0;
}
```