#include "bits/stdc++.h"

using namespace std;

typedef long long ll;
int n,m;

vector<vector<int> > calc(vector<vector<int> > mat,ll t){
    if(t==1) return mat;
	vector<vector<int> > tmp=calc(mat,t/2),res(n,vector<int>(n,0));
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			for(int k=0;k<n;k++){
				res[i][j]+=tmp[i][k]*tmp[k][j];
				res[i][j]%=m;
			}
		}
	}
	if(t%2==0) return res;
	tmp=res;
	for(int i=0;i<n;i++)for(int j=0;j<n;j++) res[i][j]=0;
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			for(int k=0;k<n;k++){
				res[i][j]+=tmp[i][k]*mat[k][j];
				res[i][j]%=m;
			}
		}
	}
	return res;
}

int pow(int x,ll a){
	if(a==0) return 1;
	int tmp=pow(x,a/2);
	if(a%2) return (tmp*tmp*x)%m;
	else return (tmp*tmp)%m;
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int a,b,c;ll t;
	while(cin>>n>>m>>a>>b>>c>>t,n){
		vector<int> s(n);
		for(int i=0;i<n;i++) cin>>s[i];
		if(n==1) cout<<pow(s[0],t)<<endl;
		else if(t==0){
			cout<<s[0];
			for(int i=1;i<n;i++) cout<<" "<<s[i];
			cout<<endl;
		}
		else{
			vector<vector<int> > mat(n,vector<int>(n,0));
			mat[0][0]=b;mat[0][1]=c;
			for(int i=1;i<n-1;i++){
				mat[i][i-1]=a;
				mat[i][i]=b;
				mat[i][i+1]=c;
			}
			mat[n-1][n-2]=a;mat[n-1][n-1]=b;
			vector<vector<int> > res=calc(mat,t);
            /*
            for(int i=0;i<n;i++){
				for(int j=0;j<n;j++){
					cout<<res[i][j];
				}
				cout<<endl;
			}*/
			vector<int> ans(n,0);
			for(int i=0;i<n;i++){
				for(int j=0;j<n;j++){
					ans[i]+=res[i][j]*s[j];
					ans[i]%=m;
				}
			}
			cout<<ans[0];
			for(int i=1;i<n;i++) 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 "bits/stdc++.h"

using namespace std;

typedef long long ll;
int n,m;

vector<vector<int> > calc(vector<vector<int> > mat,ll t){
    if(t==1) return mat;
	vector<vector<int> > tmp=calc(mat,t/2),res(n,vector<int>(n,0));
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			for(int k=0;k<n;k++){
				res[i][j]+=tmp[i][k]*tmp[k][j];
				res[i][j]%=m;
			}
		}
	}
	if(t%2==0) return res;
	tmp=res;
	for(int i=0;i<n;i++)for(int j=0;j<n;j++) res[i][j]=0;
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			for(int k=0;k<n;k++){
				res[i][j]+=tmp[i][k]*mat[k][j];
				res[i][j]%=m;
			}
		}
	}
	return res;
}

int pow(int x,ll a){
	if(a==0) return 1;
	int tmp=pow(x,a/2);
	if(a%2) return (tmp*tmp*x)%m;
	else return (tmp*tmp)%m;
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int a,b,c;ll t;
	while(cin>>n>>m>>a>>b>>c>>t,n){
		vector<int> s(n);
		for(int i=0;i<n;i++) cin>>s[i];
		if(n==1) cout<<pow(s[0],t)<<endl;
		else if(t==0){
			cout<<s[0];
			for(int i=1;i<n;i++) cout<<" "<<s[i];
			cout<<endl;
		}
		else{
			vector<vector<int> > mat(n,vector<int>(n,0));
			mat[0][0]=b;mat[0][1]=c;
			for(int i=1;i<n-1;i++){
				mat[i][i-1]=a;
				mat[i][i]=b;
				mat[i][i+1]=c;
			}
			mat[n-1][n-2]=a;mat[n-1][n-1]=b;
			vector<vector<int> > res=calc(mat,t);
            /*
            for(int i=0;i<n;i++){
				for(int j=0;j<n;j++){
					cout<<res[i][j];
				}
				cout<<endl;
			}*/
			vector<int> ans(n,0);
			for(int i=0;i<n;i++){
				for(int j=0;j<n;j++){
					ans[i]+=res[i][j]*s[j];
					ans[i]%=m;
				}
			}
			cout<<ans[0];
			for(int i=1;i<n;i++) cout<<" "<<ans[i];
			cout<<endl;
		}
	}
}


```