#include <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <climits>
#include <cfloat>
using namespace std;

int MOD;

template <class T>
vector<vector<T> > matrixProduct(const vector<vector<T> >& x, const vector<vector<T> >& y)
{
    int a = x.size();
    int b = x[0].size();
    int c = y[0].size();
    vector<vector<T> > z(a, vector<T>(c, 0));
    for(int i=0; i<a; ++i){
        for(int j=0; j<c; ++j){
            for(int k=0; k<b; ++k){
                z[i][j] += x[i][k] * y[k][j];
                z[i][j] %= MOD;
            }
        }
    }
    return z;
}

template <class T>
vector<vector<T> > matrixPower(const vector<vector<T> >& x, int k)
{
    int n = x.size();
    vector<vector<T> > y(n, vector<T>(n, 0));
    for(int i=0; i<n; ++i)
        y[i][i] = 1;

    vector<vector<T> > z = x;
    while(k > 0){
        if(k & 1)
            y = matrixProduct(y, z);
        z = matrixProduct(z, z);
        k >>= 1;
    }
    return y;
}

int main()
{
    for(;;){
        int n, a, b, c, t;
        cin >> n >> MOD >> a >> b >> c >> t;
        if(n == 0)
            return 0;

        vector<vector<int> > s(n, vector<int>(1));
        for(int i=0; i<n; ++i)
            cin >> s[i][0];

        vector<vector<int> > mat(n, vector<int>(n, 0));
        for(int i=0; i<n; ++i){
            mat[i][i] = b;
            if(i-1 >= 0)
                mat[i][i-1] = a;
            if(i+1 < n)
                mat[i][i+1] = c;
        }
        mat = matrixPower(mat, t);

        vector<vector<int> > ret = matrixProduct(mat, s);
        cout << ret[0][0];
        for(int i=1; i<n; ++i)
            cout << ' ' << ret[i][0];
        cout << endl;
    }
}
### Prompt

Create a solution in CPP for 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 <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <climits>
#include <cfloat>
using namespace std;

int MOD;

template <class T>
vector<vector<T> > matrixProduct(const vector<vector<T> >& x, const vector<vector<T> >& y)
{
    int a = x.size();
    int b = x[0].size();
    int c = y[0].size();
    vector<vector<T> > z(a, vector<T>(c, 0));
    for(int i=0; i<a; ++i){
        for(int j=0; j<c; ++j){
            for(int k=0; k<b; ++k){
                z[i][j] += x[i][k] * y[k][j];
                z[i][j] %= MOD;
            }
        }
    }
    return z;
}

template <class T>
vector<vector<T> > matrixPower(const vector<vector<T> >& x, int k)
{
    int n = x.size();
    vector<vector<T> > y(n, vector<T>(n, 0));
    for(int i=0; i<n; ++i)
        y[i][i] = 1;

    vector<vector<T> > z = x;
    while(k > 0){
        if(k & 1)
            y = matrixProduct(y, z);
        z = matrixProduct(z, z);
        k >>= 1;
    }
    return y;
}

int main()
{
    for(;;){
        int n, a, b, c, t;
        cin >> n >> MOD >> a >> b >> c >> t;
        if(n == 0)
            return 0;

        vector<vector<int> > s(n, vector<int>(1));
        for(int i=0; i<n; ++i)
            cin >> s[i][0];

        vector<vector<int> > mat(n, vector<int>(n, 0));
        for(int i=0; i<n; ++i){
            mat[i][i] = b;
            if(i-1 >= 0)
                mat[i][i-1] = a;
            if(i+1 < n)
                mat[i][i+1] = c;
        }
        mat = matrixPower(mat, t);

        vector<vector<int> > ret = matrixProduct(mat, s);
        cout << ret[0][0];
        for(int i=1; i<n; ++i)
            cout << ' ' << ret[i][0];
        cout << endl;
    }
}
```