#include<bits/stdc++.h>
#define int long long
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

int M;

template<class T> class Matrix
{
 public:
    vector<vector<T>> A;
    Matrix(int n):A(n,vector<T>(n)){}
    Matrix(int n,int m):A(n,vector<T>(m)){}

    vector<T>& operator[](int i){ return A[i]; }
    int height(){ return A.size(); }
    int width(){ return A[0].size(); }
    static Matrix I(int n){
        Matrix mat(n);
        for(int i=0;i<n;i++) mat[i][i]=1;
        return mat;
    }
    Matrix& operator*=(Matrix& B){
        int H=height(), W=B.width(), K=width();
        vector<vector<T>> C(H,vector<T>(W));
        for(int i=0;i<H;i++) for(int j=0;j<W;j++)
            for(int k=0;k<K;k++) C[i][j]+=A[i][k]*B[k][j],C[i][j]%=M;
        A.swap(C);
        return *this;
    }
    Matrix operator*(Matrix& B){ return Matrix(*this)*=B; }
    Matrix pow(int K){
        Matrix res=Matrix::I(height());
        Matrix B=*this;
        while(K>0){
            if(K&1) res*=B; B*=B; K>>=1;
        }
        return res;
    }
};

void solve()
{
    int N,A,B,C,T; cin>>N>>M>>A>>B>>C>>T;
    if(N==0) exit(0);
    Matrix<int> a(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;
    }
    a=a.pow(T);
    Matrix<int> b(N,1);
    for(int i=0;i<N;i++){
        cin>>b[i][0];
    }
    a*=b;
    for(int i=0;i<N;i++){
        cout<<a[i][0]<<(i==N-1? '\n':' ');   
    }
}
signed main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    while(true){
        solve();
    }
}

### 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<bits/stdc++.h>
#define int long long
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

int M;

template<class T> class Matrix
{
 public:
    vector<vector<T>> A;
    Matrix(int n):A(n,vector<T>(n)){}
    Matrix(int n,int m):A(n,vector<T>(m)){}

    vector<T>& operator[](int i){ return A[i]; }
    int height(){ return A.size(); }
    int width(){ return A[0].size(); }
    static Matrix I(int n){
        Matrix mat(n);
        for(int i=0;i<n;i++) mat[i][i]=1;
        return mat;
    }
    Matrix& operator*=(Matrix& B){
        int H=height(), W=B.width(), K=width();
        vector<vector<T>> C(H,vector<T>(W));
        for(int i=0;i<H;i++) for(int j=0;j<W;j++)
            for(int k=0;k<K;k++) C[i][j]+=A[i][k]*B[k][j],C[i][j]%=M;
        A.swap(C);
        return *this;
    }
    Matrix operator*(Matrix& B){ return Matrix(*this)*=B; }
    Matrix pow(int K){
        Matrix res=Matrix::I(height());
        Matrix B=*this;
        while(K>0){
            if(K&1) res*=B; B*=B; K>>=1;
        }
        return res;
    }
};

void solve()
{
    int N,A,B,C,T; cin>>N>>M>>A>>B>>C>>T;
    if(N==0) exit(0);
    Matrix<int> a(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;
    }
    a=a.pow(T);
    Matrix<int> b(N,1);
    for(int i=0;i<N;i++){
        cin>>b[i][0];
    }
    a*=b;
    for(int i=0;i<N;i++){
        cout<<a[i][0]<<(i==N-1? '\n':' ');   
    }
}
signed main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    while(true){
        solve();
    }
}

```