#include <bits/stdc++.h>
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
int M;

void mul(vvi &A, const vvi B) {
    vvi t(A.size(), vi(B[0].size(), 0));
    for (size_t i=0; i<A.size(); ++i)
        for (size_t j=0; j<B[0].size(); ++j)
            for (size_t k=0; k<A[0].size(); ++k)
                (t[i][j] += (A[i][k] * B[k][j]) % M) %= M;
    swap(A, t);
}

void pow(vvi &A, int k) {
    int n = A.size();
    vvi t(n, vi(n, 0));
    for (int i=0; i<n; ++i) t[i][i] = 1;
    while (k) {
        if (k & 1) mul(t, A);
        k >>= 1;
        mul(A, A);
    }
    swap(A, t);
}

int main() {
    int n, a, b, c, t;
    while (cin >> n >> M >> a >> b >> c >> t) {
        if ( ! n) break;
        vvi A(n, vi(n, 0));
        for (int i=0; i<n; ++i) cin >> A[0][i];

        vvi B(n, vi(n, 0));
        for (int i=0; i<n; ++i) {
            if (i+1 < n) B[i][i+1] = a;
            B[i][i] = b;
            if (0 <= i-1) B[i][i-1] = c;
        }
        pow(B, t);
        mul(A, B);
        for (int i=0; i<n; ++i) cout << (i==0 ? "" : " ") << A[0][i];
        cout << endl;
    }
}

### Prompt

Develop a solution in cpp 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 <bits/stdc++.h>
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
int M;

void mul(vvi &A, const vvi B) {
    vvi t(A.size(), vi(B[0].size(), 0));
    for (size_t i=0; i<A.size(); ++i)
        for (size_t j=0; j<B[0].size(); ++j)
            for (size_t k=0; k<A[0].size(); ++k)
                (t[i][j] += (A[i][k] * B[k][j]) % M) %= M;
    swap(A, t);
}

void pow(vvi &A, int k) {
    int n = A.size();
    vvi t(n, vi(n, 0));
    for (int i=0; i<n; ++i) t[i][i] = 1;
    while (k) {
        if (k & 1) mul(t, A);
        k >>= 1;
        mul(A, A);
    }
    swap(A, t);
}

int main() {
    int n, a, b, c, t;
    while (cin >> n >> M >> a >> b >> c >> t) {
        if ( ! n) break;
        vvi A(n, vi(n, 0));
        for (int i=0; i<n; ++i) cin >> A[0][i];

        vvi B(n, vi(n, 0));
        for (int i=0; i<n; ++i) {
            if (i+1 < n) B[i][i+1] = a;
            B[i][i] = b;
            if (0 <= i-1) B[i][i-1] = c;
        }
        pow(B, t);
        mul(A, B);
        for (int i=0; i<n; ++i) cout << (i==0 ? "" : " ") << A[0][i];
        cout << endl;
    }
}

```