#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;
inline ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
inline ll lcm(ll a, ll b) { return a / gcd(a, b)*b; }

typedef vector<ll> vec;
typedef vector<vec> mat;
/*
 行列累乗
 X = A^M*B
 A ( N*N行列)
 
 O(N^3 logM)
 */
mat mul(mat&A,mat&B,const ll M){
    mat C(A.size(),vec(B[0].size()));
    for(int i = 0; i < (int)A.size();i++){
        for(int k = 0; k < (int)B.size();k++){
            if(A[i][k] == 0) continue;
            for(int j = 0; j < B[0].size();j++){
                C[i][j] = (C[i][j] + A[i][k]*B[k][j])%M;
            }
        }
    }
    return C;
}

mat pow(mat A,ll n,const ll M){
    mat B(A.size(),vec(A.size()));
    for(int i = 0; i < A.size();i++){
        B[i][i] = 1;
    }
    while(n > 0){
        if(n&1) B= mul(B,A,M);
        A = mul(A,A,M);
        n>>=1;
    }
    return B;
}

ll N,M,A,B,C,T;
mat S;
int main(void) {
    cin.tie(0); ios_base::sync_with_stdio(false);
    while(cin >> N >> M >> A >> B >> C >> T,N){
        S = mat(N,vec(1));
        for(int i = 0; i < N;i++) cin >> S[i][0];
        
        mat D(N,vec(N));
        for(int i = 0; i < N;i++){
            if(i-1>=0) D[i][i-1] = A;
            D[i][i] = B;
            if(i+1<=N-1) D[i][i+1] = C;
        }
        
        mat DT = pow(D,T,M);
        
        mat R = mul(DT,S,M);
        for(int i = 0; i < N;i++){
            cout << R[i][0];
            if(i!=N-1) cout << " ";
            else cout << endl;
        }
    }
    return 0;
}


### Prompt

Develop a solution in Cpp to the problem described below:

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;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;
inline ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
inline ll lcm(ll a, ll b) { return a / gcd(a, b)*b; }

typedef vector<ll> vec;
typedef vector<vec> mat;
/*
 行列累乗
 X = A^M*B
 A ( N*N行列)
 
 O(N^3 logM)
 */
mat mul(mat&A,mat&B,const ll M){
    mat C(A.size(),vec(B[0].size()));
    for(int i = 0; i < (int)A.size();i++){
        for(int k = 0; k < (int)B.size();k++){
            if(A[i][k] == 0) continue;
            for(int j = 0; j < B[0].size();j++){
                C[i][j] = (C[i][j] + A[i][k]*B[k][j])%M;
            }
        }
    }
    return C;
}

mat pow(mat A,ll n,const ll M){
    mat B(A.size(),vec(A.size()));
    for(int i = 0; i < A.size();i++){
        B[i][i] = 1;
    }
    while(n > 0){
        if(n&1) B= mul(B,A,M);
        A = mul(A,A,M);
        n>>=1;
    }
    return B;
}

ll N,M,A,B,C,T;
mat S;
int main(void) {
    cin.tie(0); ios_base::sync_with_stdio(false);
    while(cin >> N >> M >> A >> B >> C >> T,N){
        S = mat(N,vec(1));
        for(int i = 0; i < N;i++) cin >> S[i][0];
        
        mat D(N,vec(N));
        for(int i = 0; i < N;i++){
            if(i-1>=0) D[i][i-1] = A;
            D[i][i] = B;
            if(i+1<=N-1) D[i][i+1] = C;
        }
        
        mat DT = pow(D,T,M);
        
        mat R = mul(DT,S,M);
        for(int i = 0; i < N;i++){
            cout << R[i][0];
            if(i!=N-1) cout << " ";
            else cout << endl;
        }
    }
    return 0;
}


```