#include <iostream>
#include <vector>
#include <cassert>
using namespace std;


int M;

typedef vector<int> Vec;
typedef vector<Vec> Mat;
Mat operator*(const Mat& x, const Mat& y) {
    assert(x[0].size() == y.size());
    Mat r(x.size(), Vec(y[0].size()));
    for (size_t i = 0; i < r.size(); i++)
        for (size_t j = 0; j < r[0].size(); j++)
            for (size_t k = 0; k < y.size(); k++)
                r[i][j] = (r[i][j] + x[i][k] * y[k][j]) % M;
    return r;
}

vector<int> solve(int n, int a, int b, int c, int t, vector<int>& S) {
    // Make transition matrix
    Mat A(n, Vec(n));
    for (int i = 0; i < n; i++) {
        if (i - 1 >= 0) A[i][i-1] = a;
        A[i][i] = b;
        if (i + 1 < n) A[i][i+1] = c;
    }

    // Make initial state
    Mat x(n, Vec(1));
    for (int i = 0; i < n; i++) x[i][0] = S[i];

    // multiply t times
    while (t) {
        if (t & 1) x = A * x;
        A = A * A;
        t >>= 1;
    }
    
    // make answer
    vector<int> res(n);
    for (int i = 0; i < n; i++) res[i] = x[i][0];
    return res;
}

int main() {
    int N, A, B, C, T;
    while (true) {
        cin >> N >> M >> A >> B >> C >> T;
        if (N == 0 && M == 0 && A == 0 && B == 0 && C == 0 && T == 0)
            break;
        vector<int> S0(N);
        for (int i = 0; i < N; i++) cin >> S0[i];
        vector<int> res = solve(N, A, B, C, T, S0);
        for (size_t i = 0; i < res.size(); i++) {
            cout << res[i];
            cout << (i != res.size() - 1 ? " " : "\n");
        }
    }
}
### Prompt

Please create a solution in cpp 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
#include <iostream>
#include <vector>
#include <cassert>
using namespace std;


int M;

typedef vector<int> Vec;
typedef vector<Vec> Mat;
Mat operator*(const Mat& x, const Mat& y) {
    assert(x[0].size() == y.size());
    Mat r(x.size(), Vec(y[0].size()));
    for (size_t i = 0; i < r.size(); i++)
        for (size_t j = 0; j < r[0].size(); j++)
            for (size_t k = 0; k < y.size(); k++)
                r[i][j] = (r[i][j] + x[i][k] * y[k][j]) % M;
    return r;
}

vector<int> solve(int n, int a, int b, int c, int t, vector<int>& S) {
    // Make transition matrix
    Mat A(n, Vec(n));
    for (int i = 0; i < n; i++) {
        if (i - 1 >= 0) A[i][i-1] = a;
        A[i][i] = b;
        if (i + 1 < n) A[i][i+1] = c;
    }

    // Make initial state
    Mat x(n, Vec(1));
    for (int i = 0; i < n; i++) x[i][0] = S[i];

    // multiply t times
    while (t) {
        if (t & 1) x = A * x;
        A = A * A;
        t >>= 1;
    }
    
    // make answer
    vector<int> res(n);
    for (int i = 0; i < n; i++) res[i] = x[i][0];
    return res;
}

int main() {
    int N, A, B, C, T;
    while (true) {
        cin >> N >> M >> A >> B >> C >> T;
        if (N == 0 && M == 0 && A == 0 && B == 0 && C == 0 && T == 0)
            break;
        vector<int> S0(N);
        for (int i = 0; i < N; i++) cin >> S0[i];
        vector<int> res = solve(N, A, B, C, T, S0);
        for (size_t i = 0; i < res.size(); i++) {
            cout << res[i];
            cout << (i != res.size() - 1 ? " " : "\n");
        }
    }
}
```