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 <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <numeric>
using namespace std;
 
#define REP(i, s, e) for (int i = (s); i < (e); i++)
#define REPI(i, s, e) for (int i = (s); i <= (e); i++)
#define rep(i, n) REP(i, 0, n)
#define repi(i, n) REPI(i, 0, n)
#define ALL(v) (v).begin(), (v).end()
 
#define dump(x) (cout << #x << " = " << x << endl)
#define dump2(x, y) (cout << "(" << #x << ", " << #y << ") = (" << x << ", " << y << ")" << endl)
#define dump3(x, y, z) (cout << "(" << #x << ", " << #y << ", " << #z << ") = (" << x << ", " << y << ", "<< z << ")" << endl)
 
typedef long long ll;
typedef pair<int, int> pii;

const int MAX = 55;

int N, M, A, B, C, T;

typedef vector< vector<int> > mat;
mat S;

mat mul(mat &A, mat &B)
{
	mat C(A.size(), vector<int>(B[0].size()));

	for (int i = 0; i < A.size(); i++) {
		for (int k = 0; k < B.size(); k++)
			for (int j = 0; j < B[0].size(); j++)
				C[i][j] = (C[i][j] + A[i][k]*B[k][j]) % M;
	}

	return C;
}

mat pow(mat A, int n)
{
	int N = A.size();
	mat B(N, vector<int>(N));

	for (int i = 0; i < N; i++)
		B[i][i] = 1;

	while (n > 0) {
		if (n&1) B = mul(B, A);
		A = mul(A, A);
		n >>= 1;
	}

	return B;
}

int main()
{
	while (cin >> N >> M >> A >> B >> C >> T, N|M|A|B|C|T) {
		S = mat(N, vector<int>(1, 0));
		rep(i, N)
			cin >> S[i][0];

		mat X = mat(N, vector<int>(N, 0));
		X[0][0] = B;
		X[0][1] = C;
		for (int i = 1; i < N-1; i++) {
			X[i][i-1] = A;
			X[i][i]   = B;
			X[i][i+1] = C;
		}
		X[N-1][N-2] = A;
		X[N-1][N-1] = B;

		X = pow(X, T);

		mat Y = mul(X, S);
		rep(i, Y.size())
			printf("%d%c", Y[i][0], (i==Y.size()-1?'\n':' '));
	}
	return 0;
}