#include <cstdio>
#include <vector>
#include <algorithm>

#define rep(i, n) for(int i = 0; i < (n); ++i)

using namespace std;

typedef vector<int> V;
typedef vector<V> M;

int n, m;
int p[3];
int t;

M mul(const M& A, const M& B){
	M C(n, V(n, 0));
	rep(i, n){
		rep(j, n){
			rep(k, n){
				C[i][j] += A[i][k] * B[k][j];
			}
			C[i][j] %= m;
		}
	}
	return C;
}

V mul(const M& A, const V& b){
	V c(n, 0);
	rep(i, n){
		rep(j, n){
			c[i] += A[i][j] * b[j];
		}
		c[i] %= m;
	}
	return c;
}

M pow(const M& A, int k){
	M B(n, V(n, 0));
	rep(i, n){
		B[i][i] = 1;
	}
	M P = A;
	while(k > 0){
		if(k & 1){
			B = mul(B, P);
		}
		P = mul(P, P);
		k >>= 1;
	}
	return B;
}

int main(){
	while(scanf("%d%d%d%d%d%d", &n, &m, &p[0], &p[1], &p[2], &t), n){
		M A(n, V(n, 0));
		rep(i, n){
			for(int j = i - 1; j <= i + 1; ++j){
				if(0 <= j && j < n){
					A[i][j] = p[j - i + 1];
				}
			}
		}
		V b(n);
		rep(i, n){
			scanf("%d", &b[i]);
		}
		A = pow(A, t);
		b = mul(A, b);
		rep(i, n){
			if(i != 0){
				putchar(' ');
			}
			printf("%d", b[i]);
		}
		putchar('\n');
	}
	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 <cstdio>
#include <vector>
#include <algorithm>

#define rep(i, n) for(int i = 0; i < (n); ++i)

using namespace std;

typedef vector<int> V;
typedef vector<V> M;

int n, m;
int p[3];
int t;

M mul(const M& A, const M& B){
	M C(n, V(n, 0));
	rep(i, n){
		rep(j, n){
			rep(k, n){
				C[i][j] += A[i][k] * B[k][j];
			}
			C[i][j] %= m;
		}
	}
	return C;
}

V mul(const M& A, const V& b){
	V c(n, 0);
	rep(i, n){
		rep(j, n){
			c[i] += A[i][j] * b[j];
		}
		c[i] %= m;
	}
	return c;
}

M pow(const M& A, int k){
	M B(n, V(n, 0));
	rep(i, n){
		B[i][i] = 1;
	}
	M P = A;
	while(k > 0){
		if(k & 1){
			B = mul(B, P);
		}
		P = mul(P, P);
		k >>= 1;
	}
	return B;
}

int main(){
	while(scanf("%d%d%d%d%d%d", &n, &m, &p[0], &p[1], &p[2], &t), n){
		M A(n, V(n, 0));
		rep(i, n){
			for(int j = i - 1; j <= i + 1; ++j){
				if(0 <= j && j < n){
					A[i][j] = p[j - i + 1];
				}
			}
		}
		V b(n);
		rep(i, n){
			scanf("%d", &b[i]);
		}
		A = pow(A, t);
		b = mul(A, b);
		rep(i, n){
			if(i != 0){
				putchar(' ');
			}
			printf("%d", b[i]);
		}
		putchar('\n');
	}
	return 0;
}
```