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
#define _USE_MATH_DEFINES
#include <algorithm>
#include <cstdio>
#include <functional>
#include <iostream>
#include <cfloat>
#include <climits>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <time.h>
#include <vector>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> i_i;
typedef pair<ll, int> ll_i;
typedef pair<double, int> d_i;
typedef pair<ll, ll> ll_ll;
typedef pair<double, double> d_d;
struct edge { int u, v; double w; };

ll MOD = 1000000007;
ll _MOD = 1000000009;
double EPS = 1e-10;

int M;

vector< vector<ll> > matmul(vector< vector<ll> >& A, vector< vector<ll> >& B) {
	int n = A.size();
	vector< vector<ll> > C(n, vector<ll>(n));
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			for (int k = 0; k < n; k++)
				C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;
	return C;
}

vector< vector<ll> > matpow(vector< vector<ll> > A, ll x) {
	int n = A.size();
	vector< vector<ll> > B(n, vector<ll>(n));
	for (int i = 0; i < n; i++) B[i][i] = 1;
	while (x > 0) {
		if (x % 2) B = matmul(B, A);
		x /= 2;
		A = matmul(A, A);
	}
	return B;
}

int main() {
	for (;;) {
		int N, A, B, C, T;
		cin >> N >> M >> A >> B >> C >> T;
		if (N == 0) break;
		vector< vector<ll> > a(N, vector<ll>(N));
		for (int i = 0; i < N; i++) {
			if (i - 1 >= 0) a[i][i - 1] = A;
			a[i][i] = B;
			if (i + 1 < N) a[i][i + 1] = C;
		}
		a = matpow(a, T);
		vector<int> S(N);
		for (int i = 0; i < N; i++) cin >> S[i];
		for (int i = 0; i < N; i++) {
			int ans = 0;
			for (int j = 0; j < N; j++) 
				ans = (ans + a[i][j] * S[j]) % M;
			cout << ans << (i + 1 < N ? ' ' : '\n');
		}
	}
}