#include <bits/stdc++.h>

using namespace std;
using i64 = long long int;
using ui64 = unsigned long long int;
using vi = vector<int>;
using vvi = vector<vi>;

////////////////////////////////

vvi mul(vvi a, vvi b, int m){
    assert(a[0].size() == b.size());
    vvi c(a.size(), vi(b[0].size()));
    for(int i=0;i<(int)a.size();i++){
        for(int j=0;j<(int)b[0].size();j++){
            for(int k=0;k<(int)b.size();k++){
                c[i][j] += a[i][k] * b[k][j];
                c[i][j] %= m;
            }
        }
    }
    return c;
}

void E(vvi &a){
    a.assign(a.size(), vi(a[0].size(), 0));
    for(int i=0;i<(int)a.size();i++)a[i][i] = 1;
}

vvi rp(const vvi &a, int n, const int &m){
    vvi r(a.size(), vi(a.size()));
    if(!n){
        E(r);
        return r;
    }
    r = rp(a, n >> 1, m);
    r = mul(r, r, m);
    if(n&1){
        return mul(a, r, m);
    }
    return r;
}

int main(){
    int n, m, a, b, c, t;
    while(cin >> n >> m >> a >> b >> c >> t, n|m|a|b|c|t){
        vvi d(n, vi(n, 0));
        for(int i=1;i<n-1;i++){
            d[i][i-1] = a;
            d[i][i] = b;
            d[i][i+1] = c;
        }
        d[0][0] = b;d[0][1] = c;d[n-1][n-2] = a;d[n-1][n-1] = b;
        
        vi e(n);
        for(auto&& x:e)cin >> x;
        vvi A = rp(d, t, m);
        
        vi ret(n, 0);
        for(int i=0; i<n; i++)
            for(int j=0;j<n;j++){
                ret[i] += A[i][j] * e[j];
                ret[i] %= m;
            }
        for(int i=0;i<n;i++)cout << ret[i] << (i==n-1 ? "\n" : " ");
    }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
using i64 = long long int;
using ui64 = unsigned long long int;
using vi = vector<int>;
using vvi = vector<vi>;

////////////////////////////////

vvi mul(vvi a, vvi b, int m){
    assert(a[0].size() == b.size());
    vvi c(a.size(), vi(b[0].size()));
    for(int i=0;i<(int)a.size();i++){
        for(int j=0;j<(int)b[0].size();j++){
            for(int k=0;k<(int)b.size();k++){
                c[i][j] += a[i][k] * b[k][j];
                c[i][j] %= m;
            }
        }
    }
    return c;
}

void E(vvi &a){
    a.assign(a.size(), vi(a[0].size(), 0));
    for(int i=0;i<(int)a.size();i++)a[i][i] = 1;
}

vvi rp(const vvi &a, int n, const int &m){
    vvi r(a.size(), vi(a.size()));
    if(!n){
        E(r);
        return r;
    }
    r = rp(a, n >> 1, m);
    r = mul(r, r, m);
    if(n&1){
        return mul(a, r, m);
    }
    return r;
}

int main(){
    int n, m, a, b, c, t;
    while(cin >> n >> m >> a >> b >> c >> t, n|m|a|b|c|t){
        vvi d(n, vi(n, 0));
        for(int i=1;i<n-1;i++){
            d[i][i-1] = a;
            d[i][i] = b;
            d[i][i+1] = c;
        }
        d[0][0] = b;d[0][1] = c;d[n-1][n-2] = a;d[n-1][n-1] = b;
        
        vi e(n);
        for(auto&& x:e)cin >> x;
        vvi A = rp(d, t, m);
        
        vi ret(n, 0);
        for(int i=0; i<n; i++)
            for(int j=0;j<n;j++){
                ret[i] += A[i][j] * e[j];
                ret[i] %= m;
            }
        for(int i=0;i<n;i++)cout << ret[i] << (i==n-1 ? "\n" : " ");
    }
}

```