#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define pi acos(-1.0)
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

VVI matmul(VVI A, VVI B, int mod){
	int n = A.size();
	VVI C(n, VI(n));
	REP(i,n) REP(j,n) REP(k,n){
		C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;
	}
	return C;
}

VVI matpow(VVI A, int t, int mod){
	int n = A.size();
	VVI B(n, VI(n));
	REP(i,n) B[i][i] = 1;
	while (t){
		if (t & 1)
			B = matmul(A, B, mod);
		A = matmul(A, A, mod);
		t /= 2;
	}
	return B;
}

int main(){
	int n, m, a, b, c, t;
	while (cin >> n >> m >> a >> b >> c >> t && n){
		VVI A(n,VI(n));
		REP(i,n){
			A[i][i] = b;
			if (i > 0) A[i][i-1] = a;
			if (i < n-1) A[i][i+1] = c;
		}
		A = matpow(A, t, m);
		VI ans(n), s(n);
		REP(i,n) cin >> s[i];
		REP(i,n) REP(j,n){
			ans[i] = (ans[i] + A[i][j] * s[j]) % m;
		}
		REP(i,n){
			cout << ans[i];
			if (i < n-1) cout << " ";
		}
		cout << endl;
	}

    return 0;
}
### Prompt

Your task is to create 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 <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define pi acos(-1.0)
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

VVI matmul(VVI A, VVI B, int mod){
	int n = A.size();
	VVI C(n, VI(n));
	REP(i,n) REP(j,n) REP(k,n){
		C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;
	}
	return C;
}

VVI matpow(VVI A, int t, int mod){
	int n = A.size();
	VVI B(n, VI(n));
	REP(i,n) B[i][i] = 1;
	while (t){
		if (t & 1)
			B = matmul(A, B, mod);
		A = matmul(A, A, mod);
		t /= 2;
	}
	return B;
}

int main(){
	int n, m, a, b, c, t;
	while (cin >> n >> m >> a >> b >> c >> t && n){
		VVI A(n,VI(n));
		REP(i,n){
			A[i][i] = b;
			if (i > 0) A[i][i-1] = a;
			if (i < n-1) A[i][i+1] = c;
		}
		A = matpow(A, t, m);
		VI ans(n), s(n);
		REP(i,n) cin >> s[i];
		REP(i,n) REP(j,n){
			ans[i] = (ans[i] + A[i][j] * s[j]) % m;
		}
		REP(i,n){
			cout << ans[i];
			if (i < n-1) cout << " ";
		}
		cout << endl;
	}

    return 0;
}
```