#include <iostream>
#include <vector>
using namespace std;

typedef long long ll;
typedef vector<vector<ll> > matrix;

int MOD;

matrix operator*(const matrix &a, const matrix &b){
	const int n = a.size();
	matrix res(n, vector<ll>(n));
	
	for(int k=0 ; k < n ; k++ ){
		for(int i=0 ; i < n ; i++ ){
			for(int j=0 ; j < n ; j++ ){
				res[i][j] += (a[i][k] * b[k][j]) % MOD;
				res[i][j] %= MOD;
			}
		}
	}
	return res;
}

vector<ll> operator*(const matrix &a, const vector<ll> &b){
	const int n = a.size();
	vector<ll> res(n);
	
	for(int i=0 ; i < n ; i++ ){
		for(int j=0 ; j < n ; j++ ){
			res[i] += (a[i][j] * b[j]) % MOD;
			res[i] %= MOD;
		}
	}
	return res;
}

matrix mod_pow(matrix m, int t){
	const int n = m.size();
	matrix res(n, vector<ll>(n));
	
	for(int i=0 ; i < n ; i++ ) res[i][i] = 1;
	while( t ){
		if( t & 1 ) res = res  * m;
		m = m * m;
		t >>= 1;
	}
	return res;
}

int main(){
	int N, M, A, B, C;
	ll T;
	while( cin >> N >> M >> A >> B >> C >> T , N ){
		vector<ll> S(N);
		MOD = M;
		for(int i=0 ; i < N ; i++ ){
			cin >> S[i];
		}
		
		matrix a(N, vector<ll>(N));
		for(int i=0 ; i < N ; i++ ){
			if( 0 <= i-1 ) a[i][i-1] = A;
			if( i+1 < N ) a[i][i+1] = C;
			a[i][i] = B;
		}
		vector<ll> ans = mod_pow(a, T) * S;
		for(int i=0 ; i < ans.size() ; i++ ){
			if( i ) cout << " ";
			cout << ans[i];
		}
		cout << endl;
	}
}
### 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 <iostream>
#include <vector>
using namespace std;

typedef long long ll;
typedef vector<vector<ll> > matrix;

int MOD;

matrix operator*(const matrix &a, const matrix &b){
	const int n = a.size();
	matrix res(n, vector<ll>(n));
	
	for(int k=0 ; k < n ; k++ ){
		for(int i=0 ; i < n ; i++ ){
			for(int j=0 ; j < n ; j++ ){
				res[i][j] += (a[i][k] * b[k][j]) % MOD;
				res[i][j] %= MOD;
			}
		}
	}
	return res;
}

vector<ll> operator*(const matrix &a, const vector<ll> &b){
	const int n = a.size();
	vector<ll> res(n);
	
	for(int i=0 ; i < n ; i++ ){
		for(int j=0 ; j < n ; j++ ){
			res[i] += (a[i][j] * b[j]) % MOD;
			res[i] %= MOD;
		}
	}
	return res;
}

matrix mod_pow(matrix m, int t){
	const int n = m.size();
	matrix res(n, vector<ll>(n));
	
	for(int i=0 ; i < n ; i++ ) res[i][i] = 1;
	while( t ){
		if( t & 1 ) res = res  * m;
		m = m * m;
		t >>= 1;
	}
	return res;
}

int main(){
	int N, M, A, B, C;
	ll T;
	while( cin >> N >> M >> A >> B >> C >> T , N ){
		vector<ll> S(N);
		MOD = M;
		for(int i=0 ; i < N ; i++ ){
			cin >> S[i];
		}
		
		matrix a(N, vector<ll>(N));
		for(int i=0 ; i < N ; i++ ){
			if( 0 <= i-1 ) a[i][i-1] = A;
			if( i+1 < N ) a[i][i+1] = C;
			a[i][i] = B;
		}
		vector<ll> ans = mod_pow(a, T) * S;
		for(int i=0 ; i < ans.size() ; i++ ){
			if( i ) cout << " ";
			cout << ans[i];
		}
		cout << endl;
	}
}
```