#include<cstdlib>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN = (int) 50 + 10;
const int INF = (int) 0x3f3f3f3f;
typedef long long LL;

int M,N,A,B,C,T;
int s[MAXN];

struct Matrix{
    LL a[MAXN][MAXN];
}st,mp;

Matrix Matrix_Multiply(const Matrix& a,const Matrix& b,int MOD){
    Matrix c;
    memset(c.a,0,sizeof(c.a));
    for(int k=1;k<=N;k++){
        for(int i=1;i<=N;i++){
            for(int j=1;j<=N;j++){
                c.a[i][k]=(c.a[i][k]+a.a[i][j]*b.a[j][k])%MOD;
            }
        }
    }
    return c;
}

void Power(int X,int MOD){
    while(X){
        if(X&1) st=Matrix_Multiply(st,mp,MOD);
        mp=Matrix_Multiply(mp,mp,MOD);
        X>>=1;
    }
}

void work(){
    while(~scanf("%d%d%d%d%d%d",&N,&M,&A,&B,&C,&T)){
        for(int i=1;i<=N;i++) scanf("%d",&s[i]);
        memset(mp.a,0,sizeof(mp.a));
        for(int i=1;i<=N;i++){
            mp.a[i][i-1]=A;
            mp.a[i][i]=B;
            mp.a[i][i+1]=C;
        }
        for(int i=1;i<=N;i++){
            for(int j=1;j<=N;j++) st.a[i][j]=(i==j);
        }
        Power(T,M);
        for(int i=1;i<=N;i++){
            LL ans=0;
            for(int j=1;j<=N;j++){
                ans=(ans+st.a[i][j]*s[j])%M;
            }
            printf("%lld%s",ans,i==N ? "\n" : " ");
        }
    }
}

int main(){
#ifdef NNever
    freopen("data.in","r",stdin);
//    freopen("out.txt","w",stdout);
#endif // NNever
    work();
    return 0;
}









### Prompt

In cpp, your task is to solve 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<cstdlib>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN = (int) 50 + 10;
const int INF = (int) 0x3f3f3f3f;
typedef long long LL;

int M,N,A,B,C,T;
int s[MAXN];

struct Matrix{
    LL a[MAXN][MAXN];
}st,mp;

Matrix Matrix_Multiply(const Matrix& a,const Matrix& b,int MOD){
    Matrix c;
    memset(c.a,0,sizeof(c.a));
    for(int k=1;k<=N;k++){
        for(int i=1;i<=N;i++){
            for(int j=1;j<=N;j++){
                c.a[i][k]=(c.a[i][k]+a.a[i][j]*b.a[j][k])%MOD;
            }
        }
    }
    return c;
}

void Power(int X,int MOD){
    while(X){
        if(X&1) st=Matrix_Multiply(st,mp,MOD);
        mp=Matrix_Multiply(mp,mp,MOD);
        X>>=1;
    }
}

void work(){
    while(~scanf("%d%d%d%d%d%d",&N,&M,&A,&B,&C,&T)){
        for(int i=1;i<=N;i++) scanf("%d",&s[i]);
        memset(mp.a,0,sizeof(mp.a));
        for(int i=1;i<=N;i++){
            mp.a[i][i-1]=A;
            mp.a[i][i]=B;
            mp.a[i][i+1]=C;
        }
        for(int i=1;i<=N;i++){
            for(int j=1;j<=N;j++) st.a[i][j]=(i==j);
        }
        Power(T,M);
        for(int i=1;i<=N;i++){
            LL ans=0;
            for(int j=1;j<=N;j++){
                ans=(ans+st.a[i][j]*s[j])%M;
            }
            printf("%lld%s",ans,i==N ? "\n" : " ");
        }
    }
}

int main(){
#ifdef NNever
    freopen("data.in","r",stdin);
//    freopen("out.txt","w",stdout);
#endif // NNever
    work();
    return 0;
}









```