#include<iostream>
#include<vector>
#include<string>
#include<algorithm>	
#include<map>
#include<set>
#include<utility>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<cstdio>
#include<sstream>
#include<iomanip>
#include<assert.h>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
using namespace std;
//kaewasuretyuui
typedef long long ll;
typedef int Def;
typedef pair<Def,Def> pii;
typedef vector<Def> vi;
typedef vector<vi> vvi;
typedef vector<pii> vp;
typedef vector<vp> vvp;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef pair<Def,pii> pip;
typedef vector<pip>vip;
//#define mt make_tuple
//typedef tuple<double,int,double> tp;
//typedef vector<tp> vt;
template<typename A,typename B>bool cmin(A &a,const B &b){return a>b?(a=b,true):false;}
template<typename A,typename B>bool cmax(A &a,const B &b){return a<b?(a=b,true):false;}
const double PI=acos(-1);
const double EPS=1e-7;
Def inf=sizeof(Def)==sizeof(long long)?2e18:1e9;
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
int MOD;
vvi mul(vvi A,vvi B){
	vvi C(A.size(),vi(B[0].size()));
	rep(i,A.size())rep(k,B.size())rep(j,B[0].size())
		(C[i][j]+=A[i][k]*B[k][j])%=MOD;
	return C;
}
vvi pow(vvi A,ll n){
	vvi B(A.size(),vi(A.size()));
	rep(i,A.size())B[i][i]=1;
	while(n){
		if(n&1)B=mul(B,A);
		A=mul(A,A);
		n>>=1;
	}
	return B;
}
int main(){
	int n,a,b,c,t;
	while(cin>>n>>MOD>>a>>b>>c>>t,n){
		vi in(n);
		rep(i,n)cin>>in[i];
		vvi A(n,vi(n));
		rep(i,n){
			if(i)A[i][i-1]=a;
			A[i][i]=b;
			if(i<n-1)A[i][i+1]=c;
		}
		A=pow(A,t);
		rep(i,n){
			if(i)cout<<" ";
			int sum=0;
			rep(j,n)(sum+=A[i][j]*in[j])%=MOD;
			cout<<sum;
		}
		cout<<endl;
	}
}
### Prompt

Generate 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<string>
#include<algorithm>	
#include<map>
#include<set>
#include<utility>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<cstdio>
#include<sstream>
#include<iomanip>
#include<assert.h>
#define loop(i,a,b) for(int i=a;i<b;i++) 
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
using namespace std;
//kaewasuretyuui
typedef long long ll;
typedef int Def;
typedef pair<Def,Def> pii;
typedef vector<Def> vi;
typedef vector<vi> vvi;
typedef vector<pii> vp;
typedef vector<vp> vvp;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef pair<Def,pii> pip;
typedef vector<pip>vip;
//#define mt make_tuple
//typedef tuple<double,int,double> tp;
//typedef vector<tp> vt;
template<typename A,typename B>bool cmin(A &a,const B &b){return a>b?(a=b,true):false;}
template<typename A,typename B>bool cmax(A &a,const B &b){return a<b?(a=b,true):false;}
const double PI=acos(-1);
const double EPS=1e-7;
Def inf=sizeof(Def)==sizeof(long long)?2e18:1e9;
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
int MOD;
vvi mul(vvi A,vvi B){
	vvi C(A.size(),vi(B[0].size()));
	rep(i,A.size())rep(k,B.size())rep(j,B[0].size())
		(C[i][j]+=A[i][k]*B[k][j])%=MOD;
	return C;
}
vvi pow(vvi A,ll n){
	vvi B(A.size(),vi(A.size()));
	rep(i,A.size())B[i][i]=1;
	while(n){
		if(n&1)B=mul(B,A);
		A=mul(A,A);
		n>>=1;
	}
	return B;
}
int main(){
	int n,a,b,c,t;
	while(cin>>n>>MOD>>a>>b>>c>>t,n){
		vi in(n);
		rep(i,n)cin>>in[i];
		vvi A(n,vi(n));
		rep(i,n){
			if(i)A[i][i-1]=a;
			A[i][i]=b;
			if(i<n-1)A[i][i+1]=c;
		}
		A=pow(A,t);
		rep(i,n){
			if(i)cout<<" ";
			int sum=0;
			rep(j,n)(sum+=A[i][j]*in[j])%=MOD;
			cout<<sum;
		}
		cout<<endl;
	}
}
```