#include <vector>
#include <list>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <utility>
#include <functional>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <climits>
#include <cassert>
using namespace std;
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
const double EPS = 1e-9;
const double PI  = acos(-1.0);

vvi mul(vvi &m,vvi &n,int mod){
	int sz=m.size();
	vvi ret(sz,vi(sz));
	REP(i,sz){
		REP(j,sz){
			REP(k,sz){
				ret[i][j]+=m[i][k]*n[k][j];
				ret[i][j]=ret[i][j]%mod;
			}
		}
	}
	return ret;
}
vvi pow(vvi m,int t,int mod){
	int sz=m.size();
	vvi ret(sz,vi(sz,0));
	REP(i,sz){
		ret[i][i]=1;
	}
	while(t){
		if(t&1){
			ret=mul(ret,m,mod);
		}
		m=mul(m,m,mod);
		t>>=1;
	}
	return ret;
}
int main(){
	int n,m,a,b,c,t;
	while(cin>>n>>m>>a>>b>>c>>t,n|m|a|b|c|t){
		vi s(n);
		REP(i,n){
			cin>>s[i];
		}
		vvi matrix(n,vi(n));
		REP(i,n){
			FOR(j,i-1,i+1+1){
				if(j>=0&&j<n){
					int abc[3]={a,b,c};
					matrix[i][j]=abc[j-(i-1)];
				}
			}
		}
		vvi powed_matrix=pow(matrix,t,m);
		vvi ans(n,vi(n,0));
		REP(i,n){
			ans[i][0]=s[i];
		}
		ans=mul(powed_matrix,ans,m);
		REP(i,n){
			cout<<(i!=0?" ":"")<<ans[i][0];
		}
		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 <vector>
#include <list>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <utility>
#include <functional>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <climits>
#include <cassert>
using namespace std;
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
const double EPS = 1e-9;
const double PI  = acos(-1.0);

vvi mul(vvi &m,vvi &n,int mod){
	int sz=m.size();
	vvi ret(sz,vi(sz));
	REP(i,sz){
		REP(j,sz){
			REP(k,sz){
				ret[i][j]+=m[i][k]*n[k][j];
				ret[i][j]=ret[i][j]%mod;
			}
		}
	}
	return ret;
}
vvi pow(vvi m,int t,int mod){
	int sz=m.size();
	vvi ret(sz,vi(sz,0));
	REP(i,sz){
		ret[i][i]=1;
	}
	while(t){
		if(t&1){
			ret=mul(ret,m,mod);
		}
		m=mul(m,m,mod);
		t>>=1;
	}
	return ret;
}
int main(){
	int n,m,a,b,c,t;
	while(cin>>n>>m>>a>>b>>c>>t,n|m|a|b|c|t){
		vi s(n);
		REP(i,n){
			cin>>s[i];
		}
		vvi matrix(n,vi(n));
		REP(i,n){
			FOR(j,i-1,i+1+1){
				if(j>=0&&j<n){
					int abc[3]={a,b,c};
					matrix[i][j]=abc[j-(i-1)];
				}
			}
		}
		vvi powed_matrix=pow(matrix,t,m);
		vvi ans(n,vi(n,0));
		REP(i,n){
			ans[i][0]=s[i];
		}
		ans=mul(powed_matrix,ans,m);
		REP(i,n){
			cout<<(i!=0?" ":"")<<ans[i][0];
		}
		cout<<endl;
	}
}
```