#include <iostream>
#include <vector>

using namespace std;

vector<vector<int> > matrix_pow(vector<vector<int> > a, vector<vector<int> > b, int m) {
    int len = a[0].size();
    vector<vector<int> > c(len, vector<int>(len, 0));
    
    for (int i = 0; i < len; i++) {
        for (int j = 0; j < len; j++) {
            for (int k = 0; k < len; k++) {
                c[i][j] += a[i][k] * b[k][j];
            }
        }
    } 
    for (int i = 0; i < len; i++) {
        for (int j = 0; j < len; j++) {
            c[i][j] %= m;
        }
    }
    return c;
}

int main(void){
    int n,m,a,b,c,t;
    while (cin>>n>>m>>a>>b>>c>>t && n) {
        vector<vector<int> > s(n, vector<int>(n, 0));
        vector<vector<int> > d = s, e = s;
        for (int i = 0; i < n; i++) cin >> s[i][0];
        
        for (int i = 0; i < n; i++) {
            if (i != 0) d[i][i-1] = a;
            d[i][i] = b;
            if (i != n-1) d[i][i+1] = c;
        }
        
        for (int i = 0; i < n; i++) e[i][i] = 1;
        while (t) {
            if (t & 1) e = matrix_pow(e, d, m);
            d = matrix_pow(d, d, m);
            t >>= 1;
        }
        s = matrix_pow(e, s, m);
        for (int i = 0; i < n; i++) cout << s[i][0] << (i==n-1 ? "\n" : " ");
    }
}
### Prompt

In Cpp, your task is to solve 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 <iostream>
#include <vector>

using namespace std;

vector<vector<int> > matrix_pow(vector<vector<int> > a, vector<vector<int> > b, int m) {
    int len = a[0].size();
    vector<vector<int> > c(len, vector<int>(len, 0));
    
    for (int i = 0; i < len; i++) {
        for (int j = 0; j < len; j++) {
            for (int k = 0; k < len; k++) {
                c[i][j] += a[i][k] * b[k][j];
            }
        }
    } 
    for (int i = 0; i < len; i++) {
        for (int j = 0; j < len; j++) {
            c[i][j] %= m;
        }
    }
    return c;
}

int main(void){
    int n,m,a,b,c,t;
    while (cin>>n>>m>>a>>b>>c>>t && n) {
        vector<vector<int> > s(n, vector<int>(n, 0));
        vector<vector<int> > d = s, e = s;
        for (int i = 0; i < n; i++) cin >> s[i][0];
        
        for (int i = 0; i < n; i++) {
            if (i != 0) d[i][i-1] = a;
            d[i][i] = b;
            if (i != n-1) d[i][i+1] = c;
        }
        
        for (int i = 0; i < n; i++) e[i][i] = 1;
        while (t) {
            if (t & 1) e = matrix_pow(e, d, m);
            d = matrix_pow(d, d, m);
            t >>= 1;
        }
        s = matrix_pow(e, s, m);
        for (int i = 0; i < n; i++) cout << s[i][0] << (i==n-1 ? "\n" : " ");
    }
}
```