#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vv;
int N,M,A,B,C,T;
inline void add(int& x,int y){
	x=(x+y)%M;
}
vv pro(vv a,vv b){
	vv c=vv(N,vi(N,0));
	rep(i,N) rep(j,N) rep(k,N) add(c[i][j],a[i][k]*b[k][j]);
	return c;
}
vv ex(vv a,int p){
	vv x=vv(N,vi(N,0));
	rep(i,N) x[i][i]=1;
	while(p){
		if(p%2) x=pro(x,a);
		a=pro(a,a);
		p/=2;
	}
	return x;
}
int s[50];
int main(){
	while(true){
		cin>>N>>M>>A>>B>>C>>T;
		if(N==0) break;
		vv c(N,vi(N,0));
		rep(i,N){
			if(i>0) c[i][i-1]=A;
			c[i][i]=B;
			if(i<N-1) c[i][i+1]=C;
		}
		c=ex(c,T);
		rep(i,N) cin>>s[i];
		rep(i,N){
			int a=0;
			rep(j,N) a+=s[j]*c[i][j];
			cout<<a%M<<(i==N-1?"\n":" ");
		}
	}
}
### 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>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vv;
int N,M,A,B,C,T;
inline void add(int& x,int y){
	x=(x+y)%M;
}
vv pro(vv a,vv b){
	vv c=vv(N,vi(N,0));
	rep(i,N) rep(j,N) rep(k,N) add(c[i][j],a[i][k]*b[k][j]);
	return c;
}
vv ex(vv a,int p){
	vv x=vv(N,vi(N,0));
	rep(i,N) x[i][i]=1;
	while(p){
		if(p%2) x=pro(x,a);
		a=pro(a,a);
		p/=2;
	}
	return x;
}
int s[50];
int main(){
	while(true){
		cin>>N>>M>>A>>B>>C>>T;
		if(N==0) break;
		vv c(N,vi(N,0));
		rep(i,N){
			if(i>0) c[i][i-1]=A;
			c[i][i]=B;
			if(i<N-1) c[i][i+1]=C;
		}
		c=ex(c,T);
		rep(i,N) cin>>s[i];
		rep(i,N){
			int a=0;
			rep(j,N) a+=s[j]*c[i][j];
			cout<<a%M<<(i==N-1?"\n":" ");
		}
	}
}
```