#include<bits/stdc++.h>
using namespace std;

#define int long long

typedef pair<int,int>pint;
typedef vector<int>vint;
typedef vector<pint>vpint;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,f,n) for(int i=(f);i<(n);i++)
#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();it++)
template<class T,class U>void chmin(T &t,U f){if(t>f)t=f;}
template<class T,class U>void chmax(T &t,U f){if(t<f)t=f;}

int mod;
typedef vector<vint>mat;
mat mul(mat A,mat B){
    mat C(A.size(),vint(B[0].size()));
    rep(i,C.size()){
        rep(j,C[0].size()){
            rep(k,A[0].size()){
                (C[i][j]+=A[i][k]*B[k][j])%=mod;
            }
        }
    }
    return C;
}
mat mat_pow(mat A,int m){
    mat B(A.size(),vint(A.size()));
    rep(i,B.size())B[i][i]=1;
    while(m){
        if(m&1)B=mul(B,A);
        A=mul(A,A);
        m>>=1;
    }
    return B;
}

int N,A,B,C,T;
int S[50];
void solve(){
    rep(i,N)cin>>S[i];
    mat M(N,vint(N));
    rep(i,N){
        if(i)M[i][i-1]=A;
        M[i][i]=B;
        if(i+1<N)M[i][i+1]=C;
    }

    mat X(N,vint(1));
    rep(i,N)X[i][0]=S[i];



    X=mul(mat_pow(M,T),X);
    rep(i,N){
        if(i)cout<<" ";
        cout<<X[i][0];
    }cout<<endl;
}

signed main(){
    while(cin>>N>>mod>>A>>B>>C>>T,N||mod||A||B||C||T)solve();
    return 0;
}
### 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;

#define int long long

typedef pair<int,int>pint;
typedef vector<int>vint;
typedef vector<pint>vpint;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,f,n) for(int i=(f);i<(n);i++)
#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();it++)
template<class T,class U>void chmin(T &t,U f){if(t>f)t=f;}
template<class T,class U>void chmax(T &t,U f){if(t<f)t=f;}

int mod;
typedef vector<vint>mat;
mat mul(mat A,mat B){
    mat C(A.size(),vint(B[0].size()));
    rep(i,C.size()){
        rep(j,C[0].size()){
            rep(k,A[0].size()){
                (C[i][j]+=A[i][k]*B[k][j])%=mod;
            }
        }
    }
    return C;
}
mat mat_pow(mat A,int m){
    mat B(A.size(),vint(A.size()));
    rep(i,B.size())B[i][i]=1;
    while(m){
        if(m&1)B=mul(B,A);
        A=mul(A,A);
        m>>=1;
    }
    return B;
}

int N,A,B,C,T;
int S[50];
void solve(){
    rep(i,N)cin>>S[i];
    mat M(N,vint(N));
    rep(i,N){
        if(i)M[i][i-1]=A;
        M[i][i]=B;
        if(i+1<N)M[i][i+1]=C;
    }

    mat X(N,vint(1));
    rep(i,N)X[i][0]=S[i];



    X=mul(mat_pow(M,T),X);
    rep(i,N){
        if(i)cout<<" ";
        cout<<X[i][0];
    }cout<<endl;
}

signed main(){
    while(cin>>N>>mod>>A>>B>>C>>T,N||mod||A||B||C||T)solve();
    return 0;
}
```