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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;
typedef long long ll;

int main() {
    while (true) {
        ll n, m, a, b, c, t;
        cin >> n >> m >> a >> b >> c >> t;
        if (max({n, m, a, b, c, t}) == 0)break;

        vector<vector<vector<ll>>> edge(40, vector<vector<ll>>(n, vector<ll>(n)));
        for (int i = 0; i < n; i++) {
            if (i > 0)edge[0][i][i - 1] = c;
            edge[0][i][i] = b;
            if (i + 1 < n)edge[0][i][i + 1] = a;
        }

        for (int i = 1; i < 40; i++) {
            for (int from = 0; from < n; from++) {
                for (int mid = 0; mid < n; mid++) {
                    for (int to = 0; to < n; to++) {
                        edge[i][from][to] += edge[i - 1][from][mid] * edge[i - 1][mid][to];
                        edge[i][from][to] %= m;
                    }
                }
            }
        }

        vector<ll> dp(n);
        for (int i = 0; i < n; i++)cin >> dp[i];

        for (int i = 0; i < 40; i++) {
            if ((t >> i) & 1) {
                vector<ll> after(n);
                for (int j = 0; j < n; j++) {
                    for (int k = 0; k < n; k++) {
                        after[k] += edge[i][j][k] * dp[j];
                        after[k] %= m;
                    }
                }
                dp = after;
            }
        }

        for (int i = 0; i < n; i++) {
            if (i > 0)cout << " ";
            cout << dp[i];
        }
        cout << endl;
    }
    return 0;
}
