//Name: One-Dimensional Cellular Automaton
//Level: 
//Category:
//Note:

/**
 */
#include <iostream>
#include <vector>
#include <string>
#include <array>

using namespace std;

typedef vector<int> MatRow;
typedef vector<vector<int>> Matrix;

void matmul(Matrix &a, const Matrix &b, int N, int M) {
    Matrix tmp(N, MatRow(N, 0));
    for(int i = 0; i < N; ++i) {
        for(int j = 0; j < N; ++j) {
            for(int k = 0; k < N; ++k) {
                tmp[i][j] += a[i][k] * b[k][j];
                tmp[i][j] %= M;
            }
        }
    }
    a = tmp;
}

bool solve() {
    int N, M, A, B, C, T;
    cin >> N >> M >> A >> B >> C >> T;
    if(!N && !M && !A && !B && !C && !T) return false;
    vector<int> init(N, 0);
    for(int i = 0; i < N; ++i) {
        cin >> init[i];
    }
    Matrix mat(N, MatRow(N, 0)), unit(N, MatRow(N, 0));
    for(int i = 0; i < N; ++i) {
        if(i > 0) mat[i][i-1] = A;
        mat[i][i] = B;
        if(i+1 < N) mat[i][i+1] = C;
    }
    for(int i = 0; i < N; ++i) {
        unit[i][i] = 1;
    }
    Matrix res(unit), ord(mat);
    while(T > 0) {
        if(T & 1) {
            matmul(res, ord, N, M);
        }
        matmul(ord, ord, N, M);
        T /= 2;
    }

    for(int i = 0; i < N; ++i) {
        int s = 0;
        for(int j = 0; j < N; ++j) {
            s += res[i][j] * init[j];
            s %= M;
        }
        if(i > 0) cout << " ";
        cout << s;
    }
    cout << endl;
    return true;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);

    while(solve()) ;
    return 0;
}
### Prompt

Your challenge is to write a CPP solution to 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
//Name: One-Dimensional Cellular Automaton
//Level: 
//Category:
//Note:

/**
 */
#include <iostream>
#include <vector>
#include <string>
#include <array>

using namespace std;

typedef vector<int> MatRow;
typedef vector<vector<int>> Matrix;

void matmul(Matrix &a, const Matrix &b, int N, int M) {
    Matrix tmp(N, MatRow(N, 0));
    for(int i = 0; i < N; ++i) {
        for(int j = 0; j < N; ++j) {
            for(int k = 0; k < N; ++k) {
                tmp[i][j] += a[i][k] * b[k][j];
                tmp[i][j] %= M;
            }
        }
    }
    a = tmp;
}

bool solve() {
    int N, M, A, B, C, T;
    cin >> N >> M >> A >> B >> C >> T;
    if(!N && !M && !A && !B && !C && !T) return false;
    vector<int> init(N, 0);
    for(int i = 0; i < N; ++i) {
        cin >> init[i];
    }
    Matrix mat(N, MatRow(N, 0)), unit(N, MatRow(N, 0));
    for(int i = 0; i < N; ++i) {
        if(i > 0) mat[i][i-1] = A;
        mat[i][i] = B;
        if(i+1 < N) mat[i][i+1] = C;
    }
    for(int i = 0; i < N; ++i) {
        unit[i][i] = 1;
    }
    Matrix res(unit), ord(mat);
    while(T > 0) {
        if(T & 1) {
            matmul(res, ord, N, M);
        }
        matmul(ord, ord, N, M);
        T /= 2;
    }

    for(int i = 0; i < N; ++i) {
        int s = 0;
        for(int j = 0; j < N; ++j) {
            s += res[i][j] * init[j];
            s %= M;
        }
        if(i > 0) cout << " ";
        cout << s;
    }
    cout << endl;
    return true;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);

    while(solve()) ;
    return 0;
}
```