#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef vector<vector<ll>> V;

//行列の掛け算（正方行列）
V mult(V a,V b,int mod)
{
	V ret(a.size());
	for(int i=0;i<a.size();i++)
	{
		for(int j=0;j<a[i].size();j++)
		{
			ll tmp=0;
			for(int k=0;k<a[i].size();k++)
			{
				tmp+=a[i][k]*b[k][j];
				tmp%=mod;
			}
			ret[i].push_back(tmp);
		}
	}
	return ret;
}

//modで行列をp乗する
//p=0の場合機能しない
V ruijou(V base,ll p,int mod)
{
	V beki[40];//2べきの行列
	beki[0]=base;
	for(int i=1;i<40;i++)
		beki[i]=mult(beki[i-1],beki[i-1],mod);
	
	V ret(base.size());
	bool f=true;
	for(int i=0;i<40;i++)
	{
		ll tmp=(1LL<<(ll)i);
		if((tmp&p)>0)
		{
			if(f)
			{
				ret=beki[i];
				f=false;
			}else{
				ret=mult(ret,beki[i],mod);
			}
		}
	}
	return ret;
}

int main()
{
	while(1)
	{
		int n,m,a,b,c;
		ll t;
		cin>>n>>m>>a>>b>>c>>t;
		if(n==0)
			break;
		vector<ll> s;//元の行列
		for(int i=0;i<n;i++)
		{
			ll tmp;
			cin>>tmp;
			s.push_back(tmp);
		}
		V tt(s.size());//累乗する行列を作る
		for(int i=0;i<s.size();i++)
		{
			for(int j=0;j<s.size();j++)
				tt[i].push_back(0);
		}
		//ABCを入れる
		for(int i=0;i<s.size();i++)
		{
			if(i>0)
				tt[i-1][i]=c;
			tt[i][i]=b;
			if(i<s.size()-1)
				tt[i+1][i]=a;
		}
		if(t>0)
		{
			tt=ruijou(tt,t,m);//行列累乗
			//debug
			/*
			for(int i=0;i<tt.size();i++)
			{
				for(int j=0;j<tt[i].size();j++)
					cout<<tt[i][j]<<" ";
				cout<<endl;
			}*/
			//最後に掛け算する
			for(int i=0;i<s.size();i++)
			{
				ll ans=0;
				for(int j=0;j<s.size();j++)
				{
					ans+=tt[i][j]*s[j];
					ans%=m;
				}
				if(ans<0)
					ans+=m;
				cout<<ans;
				if(i!=s.size()-1)
					cout<<" ";
			}
			cout<<endl;
		}else{
			for(int i=0;i<s.size();i++)
			{	
				cout<<s[i];
				if(i!=s.size()-1)
					cout<<" ";
			}
			cout<<endl;
		}
	}
	return 0;
}

### Prompt

Create a solution in CPP for 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;
typedef vector<vector<ll>> V;

//行列の掛け算（正方行列）
V mult(V a,V b,int mod)
{
	V ret(a.size());
	for(int i=0;i<a.size();i++)
	{
		for(int j=0;j<a[i].size();j++)
		{
			ll tmp=0;
			for(int k=0;k<a[i].size();k++)
			{
				tmp+=a[i][k]*b[k][j];
				tmp%=mod;
			}
			ret[i].push_back(tmp);
		}
	}
	return ret;
}

//modで行列をp乗する
//p=0の場合機能しない
V ruijou(V base,ll p,int mod)
{
	V beki[40];//2べきの行列
	beki[0]=base;
	for(int i=1;i<40;i++)
		beki[i]=mult(beki[i-1],beki[i-1],mod);
	
	V ret(base.size());
	bool f=true;
	for(int i=0;i<40;i++)
	{
		ll tmp=(1LL<<(ll)i);
		if((tmp&p)>0)
		{
			if(f)
			{
				ret=beki[i];
				f=false;
			}else{
				ret=mult(ret,beki[i],mod);
			}
		}
	}
	return ret;
}

int main()
{
	while(1)
	{
		int n,m,a,b,c;
		ll t;
		cin>>n>>m>>a>>b>>c>>t;
		if(n==0)
			break;
		vector<ll> s;//元の行列
		for(int i=0;i<n;i++)
		{
			ll tmp;
			cin>>tmp;
			s.push_back(tmp);
		}
		V tt(s.size());//累乗する行列を作る
		for(int i=0;i<s.size();i++)
		{
			for(int j=0;j<s.size();j++)
				tt[i].push_back(0);
		}
		//ABCを入れる
		for(int i=0;i<s.size();i++)
		{
			if(i>0)
				tt[i-1][i]=c;
			tt[i][i]=b;
			if(i<s.size()-1)
				tt[i+1][i]=a;
		}
		if(t>0)
		{
			tt=ruijou(tt,t,m);//行列累乗
			//debug
			/*
			for(int i=0;i<tt.size();i++)
			{
				for(int j=0;j<tt[i].size();j++)
					cout<<tt[i][j]<<" ";
				cout<<endl;
			}*/
			//最後に掛け算する
			for(int i=0;i<s.size();i++)
			{
				ll ans=0;
				for(int j=0;j<s.size();j++)
				{
					ans+=tt[i][j]*s[j];
					ans%=m;
				}
				if(ans<0)
					ans+=m;
				cout<<ans;
				if(i!=s.size()-1)
					cout<<" ";
			}
			cout<<endl;
		}else{
			for(int i=0;i<s.size();i++)
			{	
				cout<<s[i];
				if(i!=s.size()-1)
					cout<<" ";
			}
			cout<<endl;
		}
	}
	return 0;
}

```