#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
typedef long long ll;

int N, M;
typedef vector<vector<int> > Mat;

Mat pow(Mat A, Mat B)
{
	vector<vector<int> > nmat(N);
	nmat.assign(N, vector<int>(N));
	
	
	for (int i=0; i<N; i++) {
		for (int j=0; j<N; j++) {
			nmat[i][j] = 0;
			for (int k=0; k<N; k++) {
				nmat[i][j] += A[i][k]*B[k][j];
				nmat[i][j] %= M;
			}
		}
	}
	
	return nmat;
}



Mat pow_mod(Mat x, ll y)
{
	vector<vector<int> > res(N); 
	res.assign(N, vector<int>(N));
	
	for (int i=0; i<N; i++) {
		for (int j=0; j<N; j++) {
			res[i][j] = i==j?1:0;
		}
	}
	
	for (; y>0; y>>=1, x=pow(x, x)) {
		if (y&1) res=pow(res, x);
	}
	
	return res;
}

int main()
{
	int  A, B, C; ll T;
	while (cin>>N>>M>>A>>B>>C>>T, N||M||A||B||C||T) {
		Mat mat(N); 
		mat.assign(N, vector<int>(N));
		
		int S[N];
		
		for (int i=0; i<N; i++) {
			cin>>S[i];
		}
		
		int base=-1;
		for (int i=0; i<N; i++) {
			for (int j=0; j<N; j++) {
				if (j==base)
					mat[i][j] = A;
				else if (j==base+1)
					mat[i][j] = B;
				else if (j==base+2)
					mat[i][j] = C;
				else
					mat[i][j] = 0;
			}
			base++;
		}
		
		mat = pow_mod(mat, T);
		// cerr << "mat = " << endl;
#if 0
		for(int i = 0; i < N; ++i) {
			for(int j = 0; j < N; ++j) {
				cerr << mat[i][j] << " ";
			}
			cerr << endl;
		}
#endif
		
		// cerr << "ans = " << endl;
		for (int i=0; i<N; i++) {
			int t=0;
			for (int j=0; j<N; j++) {
				t += S[j]*mat[i][j];
				t %= M;
			}
			if (i!=0) cout << " ";
			cout << t;
		}
		cout << endl;
	} 
}
### Prompt

Please formulate a Cpp solution to 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>
#include <cstring>
using namespace std;
typedef long long ll;

int N, M;
typedef vector<vector<int> > Mat;

Mat pow(Mat A, Mat B)
{
	vector<vector<int> > nmat(N);
	nmat.assign(N, vector<int>(N));
	
	
	for (int i=0; i<N; i++) {
		for (int j=0; j<N; j++) {
			nmat[i][j] = 0;
			for (int k=0; k<N; k++) {
				nmat[i][j] += A[i][k]*B[k][j];
				nmat[i][j] %= M;
			}
		}
	}
	
	return nmat;
}



Mat pow_mod(Mat x, ll y)
{
	vector<vector<int> > res(N); 
	res.assign(N, vector<int>(N));
	
	for (int i=0; i<N; i++) {
		for (int j=0; j<N; j++) {
			res[i][j] = i==j?1:0;
		}
	}
	
	for (; y>0; y>>=1, x=pow(x, x)) {
		if (y&1) res=pow(res, x);
	}
	
	return res;
}

int main()
{
	int  A, B, C; ll T;
	while (cin>>N>>M>>A>>B>>C>>T, N||M||A||B||C||T) {
		Mat mat(N); 
		mat.assign(N, vector<int>(N));
		
		int S[N];
		
		for (int i=0; i<N; i++) {
			cin>>S[i];
		}
		
		int base=-1;
		for (int i=0; i<N; i++) {
			for (int j=0; j<N; j++) {
				if (j==base)
					mat[i][j] = A;
				else if (j==base+1)
					mat[i][j] = B;
				else if (j==base+2)
					mat[i][j] = C;
				else
					mat[i][j] = 0;
			}
			base++;
		}
		
		mat = pow_mod(mat, T);
		// cerr << "mat = " << endl;
#if 0
		for(int i = 0; i < N; ++i) {
			for(int j = 0; j < N; ++j) {
				cerr << mat[i][j] << " ";
			}
			cerr << endl;
		}
#endif
		
		// cerr << "ans = " << endl;
		for (int i=0; i<N; i++) {
			int t=0;
			for (int j=0; j<N; j++) {
				t += S[j]*mat[i][j];
				t %= M;
			}
			if (i!=0) cout << " ";
			cout << t;
		}
		cout << endl;
	} 
}
```