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;

#define for_(i,a,b) for(int i=a;i<b;++i)
#define for_rev(i,a,b) for(int i=a;i>=b;--i)
#define allof(a) a.begin(),a.end()
#define minit(a,b) memset(a,b,sizeof(a))
#define size_of(a) (int)a.size()

typedef long long lint;
typedef pair<int, int> pii;

int N, M, A, B, C, T;
int S[55];

typedef int mat_type;
typedef vector< mat_type > mArray;
typedef vector< mArray > mMatrix;

mMatrix operator * (const mMatrix& a, const mMatrix& b) {
	int n = size_of(a), m = size_of(b[0]), b_size = size_of(b);
	mMatrix res(n, mArray(m, 0));
	for_(i,0,n) for_(k,0,b_size) for_(j,0,m) {
		res[i][j] += a[i][k] * b[k][j];
		res[i][j] %= M;
	}
	return res;
}

mMatrix mMatrixI(int n) {
	mMatrix res(n, mArray(n, 0));
	for_(i,0,n) res[i][i] = 1;
	return res;
}

mMatrix POW(mMatrix& a, lint k) {
	int n = size_of(a);
	mMatrix res = mMatrixI(n);
	
	while (k > 0) {
		if (k & 1) res = res * a;
		a = a * a;
		k >>= 1;
	}
	
	return res;
}

void solve() {
	mMatrix ABC(N, mArray(N, 0));
	int p = -1;
	
	for_(i,0,N) {
		for_(j,0,3) {
			if (p + j < 0 || p + j >= N) continue;
			if (j == 0) ABC[i][p + j] = A;
			if (j == 1) ABC[i][p + j] = B;
			if (j == 2) ABC[i][p + j] = C;
		}
		p++;
	}
	/*
	for_(i,0,N) {
		for_(j,0,N) cout << ABC[i][j] << " ";
		cout << endl;
	}*/
	
	mMatrix ABCT = POW(ABC, T);
	/*
	for_(i,0,N) {
		for_(j,0,N) cout << ABC[i][j] << " ";
		cout << endl;
	}*/
	
	mMatrix ss(N, mArray(1, 0));
	for_(i,0,N) ss[i][0] = S[i];
	
	mMatrix ans = ABCT * ss;
	
	for_(i,0,N) {
		if (i < N - 1) cout << ans[i][0] << " ";
		else cout << ans[i][0] << endl;
	}
}

int main() {
	while (cin >> N >> M >> A >> B >> C >> T, N) {
		for_(i,0,N) cin >> S[i];
		solve();
	}
	return 0;
}