#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<climits>
#include<iostream>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<numeric>

#define rep(n) for(int i=0;i<n;i++)
#define repp(j, n) for(int j=0;j<n;j++)
#define reppp(i, m, n) for(int i=m;i<=n;i++)
#define all(c) c.begin(), c.end()
#define rall(c) c.rbegin(), c.rend()
#define MOD 1000000007
#define MAX 1000000001
#define INF 1410065408
#define EPS 1e-9
#define DEBUG 0
#define ll long long
#define Pll pair<ll, ll>
#define Pii pair<int, int>
 
using namespace std;
struct edge{int from, to; ll cost;};

vector< vector<ll> > product(vector< vector<ll> > A, vector< vector<ll> > B, ll height, ll width, ll n, int m){
	
	vector< vector<ll> > C(height, vector<ll>(width, 0));
	
	rep(height){
		repp(j, width){
			repp(k, n){
				C[i][j] += A[i][k] * B[k][j];
			}
            C[i][j] %= m;
		}
	}
	
	return C;
}

signed main(){
    int n,m,t;
    ll a,b,c;
    vector< vector<ll> > ans;
    while(true){
        scanf("%d %d %lld %lld %lld %d", &n, &m, &a, &b, &c, &t);
        if(n == 0) break;

        vector< vector<ll> > s(n, vector<ll>(1));
        rep(n) scanf("%lld", &s[i][0]);

        vector< vector<ll> > A(n, vector<ll>(n, 0));
        rep(n){
            if(i > 0) A[i][i-1] = a;
            A[i][i] = b;
            if(i < n-1) A[i][i+1] = c;
        }

        vector< vector<ll> > B(n, vector<ll>(n, 0));
        rep(n) B[i][i] = 1;
        while(t){
            if(t & 1){
                B = product(B, A, n, n, n, m);
            }
            A = product(A, A, n, n, n, m);
            t = t >> 1;
        }

        vector< vector<ll> > C = product(B, s, n, 1, n, m);
        rep(n-1) printf("%lld ", C[i][0]);
        printf("%lld\n", C[n-1][0]);
    }


}
### 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<cstdlib>
#include<cmath>
#include<climits>
#include<iostream>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<numeric>

#define rep(n) for(int i=0;i<n;i++)
#define repp(j, n) for(int j=0;j<n;j++)
#define reppp(i, m, n) for(int i=m;i<=n;i++)
#define all(c) c.begin(), c.end()
#define rall(c) c.rbegin(), c.rend()
#define MOD 1000000007
#define MAX 1000000001
#define INF 1410065408
#define EPS 1e-9
#define DEBUG 0
#define ll long long
#define Pll pair<ll, ll>
#define Pii pair<int, int>
 
using namespace std;
struct edge{int from, to; ll cost;};

vector< vector<ll> > product(vector< vector<ll> > A, vector< vector<ll> > B, ll height, ll width, ll n, int m){
	
	vector< vector<ll> > C(height, vector<ll>(width, 0));
	
	rep(height){
		repp(j, width){
			repp(k, n){
				C[i][j] += A[i][k] * B[k][j];
			}
            C[i][j] %= m;
		}
	}
	
	return C;
}

signed main(){
    int n,m,t;
    ll a,b,c;
    vector< vector<ll> > ans;
    while(true){
        scanf("%d %d %lld %lld %lld %d", &n, &m, &a, &b, &c, &t);
        if(n == 0) break;

        vector< vector<ll> > s(n, vector<ll>(1));
        rep(n) scanf("%lld", &s[i][0]);

        vector< vector<ll> > A(n, vector<ll>(n, 0));
        rep(n){
            if(i > 0) A[i][i-1] = a;
            A[i][i] = b;
            if(i < n-1) A[i][i+1] = c;
        }

        vector< vector<ll> > B(n, vector<ll>(n, 0));
        rep(n) B[i][i] = 1;
        while(t){
            if(t & 1){
                B = product(B, A, n, n, n, m);
            }
            A = product(A, A, n, n, n, m);
            t = t >> 1;
        }

        vector< vector<ll> > C = product(B, s, n, 1, n, m);
        rep(n-1) printf("%lld ", C[i][0]);
        printf("%lld\n", C[n-1][0]);
    }


}
```