#include<iostream>
#include<sstream>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<complex>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>

#define rep(i,n) for(int i=0;i<(int)n;i++)
#define all(c) (c).begin(),(c).end()
#define mp make_pair
#define pb push_back
#define each(i,c) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();i++)
#define dbg(x) cerr<<__LINE__<<": "<<#x<<" = "<<(x)<<endl

using namespace std;

typedef long long ll;
typedef vector<ll> vi;
typedef pair<int,int> pi;
const int inf = (int)1e9;
const double INF = 1e12, EPS = 1e-9;


int n, mod, a, b, c, t;
typedef vector<vi> M;
inline M operator*(const M & a, const M &b){
	M c(a.size(), vi(b[0].size()));
	rep(i, c.size()) rep(j, c[0].size()){
		ll s = 0;
		rep(k, a[0].size()) s += (ll)a[i][k] * b[k][j] % mod;
		c[i][j] = s % mod;
	}
	return c;
}
inline M pow(M a, ll m){
	M res(a.size(), vi(a.size()));
	rep(i, a.size()) res[i][i] = 1;
	for(; m; m /= 2){
		if(m & 1) res = res * a;
		a = a * a;
	}
	return res;
}

int main(){
	while(cin >> n >> mod >> a >> b >> c >> t, n){
		M m(n, vi(n));
		rep(i, n){
			if(i) m[i][i - 1] = a;
			m[i][i] = b;
			if(i < n - 1) m[i][i + 1] = c;
		}
		m = pow(m, t);
		M s(n, vi(1));
		rep(i, n) cin >> s[i][0];
		s = m * s;
		
		rep(i, n) cout << s[i][0] << (i == n - 1 ? "\n" : " ");
	}
	return 0;
}
### 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<sstream>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<complex>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>

#define rep(i,n) for(int i=0;i<(int)n;i++)
#define all(c) (c).begin(),(c).end()
#define mp make_pair
#define pb push_back
#define each(i,c) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();i++)
#define dbg(x) cerr<<__LINE__<<": "<<#x<<" = "<<(x)<<endl

using namespace std;

typedef long long ll;
typedef vector<ll> vi;
typedef pair<int,int> pi;
const int inf = (int)1e9;
const double INF = 1e12, EPS = 1e-9;


int n, mod, a, b, c, t;
typedef vector<vi> M;
inline M operator*(const M & a, const M &b){
	M c(a.size(), vi(b[0].size()));
	rep(i, c.size()) rep(j, c[0].size()){
		ll s = 0;
		rep(k, a[0].size()) s += (ll)a[i][k] * b[k][j] % mod;
		c[i][j] = s % mod;
	}
	return c;
}
inline M pow(M a, ll m){
	M res(a.size(), vi(a.size()));
	rep(i, a.size()) res[i][i] = 1;
	for(; m; m /= 2){
		if(m & 1) res = res * a;
		a = a * a;
	}
	return res;
}

int main(){
	while(cin >> n >> mod >> a >> b >> c >> t, n){
		M m(n, vi(n));
		rep(i, n){
			if(i) m[i][i - 1] = a;
			m[i][i] = b;
			if(i < n - 1) m[i][i + 1] = c;
		}
		m = pow(m, t);
		M s(n, vi(1));
		rep(i, n) cin >> s[i][0];
		s = m * s;
		
		rep(i, n) cout << s[i][0] << (i == n - 1 ? "\n" : " ");
	}
	return 0;
}
```