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

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define all(a)  (a).begin(),(a).end()
#define pb push_back
#define INF (1e9+1)
//#define INF (1LL<<59)

int n,MOD,a,b,c,t;

template<class T>
class matrix{
public:
	vector<vector<T>> elm;
	int height,width;
	
	
	matrix():height(0),width(0){};
	matrix(int h,int w):height(h),width(w){
		elm.resize(h,vector<T>(w,0));	//0-padding
	}
	
	matrix<T> operator + (const matrix &m)const{
		matrix<T> ret = this;
		rep(i,height) rep(j,width) ret[i][j] += m.elm[i][j];
		return ret;
	}
	matrix<T> operator - (const matrix m)const{
		matrix<T> ret = this;
		rep(i,height) rep(j,width) ret[i][j] -= m.elm[i][j];
		return ret;
	}
	matrix<T> operator * (const matrix m)const{
		assert(width==m.height);
		matrix<T> ret(height,m.width);
		rep(i,height){
			rep(j,m.width){
				rep(k,width){
					(ret.elm[i][j] += elm[i][k]*m.elm[k][j])%=MOD;
				}
			}
		}
		return ret;
	}
	
	void output(){
		rep(i,height){
			rep(j,width){
				if(j)cout<<" ";
				cout<<elm[i][j];
			}
			cout<<endl;
		}
	}
	
	matrix<T> tern(){
		matrix<T> ret(width,height);
		rep(i,height) rep(j,width) ret.elm[j][i] = elm[i][j];
		return ret;
	}
};


vector<matrix<int>> T;

matrix<int> f(int t,matrix<int> v){
	matrix<int> ret = v;
	int c=0;
	while(t){
		if(t%2==1)ret = T[c]*ret;
		t/=2;
		c++;
	}
	return ret;
}


int main(){
	while(cin>>n>>MOD>>a>>b>>c>>t&&(n||MOD||a||b||c||t)){
		matrix<int> v(n,1);
		rep(i,n)cin>>v.elm[i][0];
		
		T.clear();
		T.resize(30,matrix<int>(n,n));
		rep(i,n)T[0].elm[i][i]=b;
		rep(i,n-1)T[0].elm[i][i+1]=c;
		rep(i,n-1)T[0].elm[i+1][i]=a;
		
		for(int i=1;i<30;i++) T[i] = T[i-1] * T[i-1];
		
		f(t,v).tern().output();
	}
}