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
#include <bits/stdc++.h>
using namespace std;

typedef long long int ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

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

mat calc(mat &X, mat &Y) {
    mat ret(X.size(), vec(Y[0].size()));

    int row = X.size();
    int colume = Y[0].size();
    if (X[0].size() != Y.size()) return ret;
    for (int i = 0; i < row; i++) {
        for (int j = 0; j < colume; j++) {
            for (int k = 0; k < X[i].size(); k++) {
                ret[i][j] += (X[i][k] * Y[k][j]) % M;
                ret[i][j] %= M;
            }
        }
    }

    return ret;
}

mat dfs(mat& X, int Times) {
    if (Times == 0) {
        mat ret(X.size(), vec(X[0].size()));
        for (int i = 0; i < X.size() && i < X[0].size(); i++) ret[i][i] = 1;
        return ret;
    }
    auto Y = dfs(X, Times / 2);
    auto YY = calc(Y, Y);
    if (Times % 2) {
        auto ret = calc(X, YY);
        return ret;
    } else {
        return YY;
    }
}

void solve() {
    mat d(N, vec(N));
    mat v(N, vec(1));
    for (auto& i : v) for (auto &j : i) cin >> j;
    for (int i = 0; i < N; i++) {
        d[i][i] = B;
        if (i + 1 < N) d[i+1][i] = A;
        if (i + 1 < N) d[i][i+1] = C;
    }

    auto d_start = dfs(d, T);
    auto ans = calc(d_start, v);

    for (int i = 0; i < N - 1; i++) cout << ans[i][0] << " "; cout << ans[N-1][0] << endl;
}

int main() {
    while (cin >> N >> M >> A >> B >> C >> T, A || B || C || N || M || T) solve();

    return 0;
}