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;

using vec = vector<int>;
using mat = vector<vec>;

int n = 0;
int mod = 1;

vec mvmul(const mat& m, const vec& v) {
    vec res(n);
    for(int i=0; i<n; i++) {
        for(int j=0; j<n; j++) {
            res[i] += m[i][j] * v[j];
        }
        res[i] %= mod;
    }
    return res;
}

mat mmul(const mat& a, const mat& b) {
    mat res(n, vec(n));
    for(int i=0; i<n; i++) {
        for(int j=0; j<n; j++) {
            for(int k=0; k<n; k++) {
                res[i][j] += a[i][k] * b[k][j];
            }
            res[i][j] %= mod;
        }
    }
    return res;
}

mat mpow(const mat& m, int t) {
    mat res(n, vec(n));
    mat a = m;
    for(int i=0; i<n; i++) res[i][i] = 1;
    while(t) {
        if(t&1) res = mmul(res, a);
        a = mmul(a, a);
        t >>= 1;
    }
    return res;
}

int main(void) {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int a, b, c, t;
    while(cin >> n >> mod >> a >> b >> c >> t, n) {
        vec v(n);
        for(int i=0; i<n; i++) cin >> v[i];
        mat m(n, vec(n));
        for(int i=0; i<n; i++) m[i][i] = b;
        for(int i=1; i<n; i++) {
            m[i][i-1] = a;
            m[i-1][i] = c;
        }
        vec res = mvmul(mpow(m, t), v);
        cout << res[0];
        for(int i=1; i<n; i++) cout << " " << res[i];
        cout << endl;
    }
    
}