#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; i++)
using namespace std;
typedef long long ll;

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    while (1)
    {
        ll n, m, a, b, c, t;
        cin >> n >> m >> a >> b >> c >> t;
        if (n == 0)
            break;
        vector<ll> S(n);
        rep(i, n) cin >> S[i];
        ll A[60][60] = {}, AA[60][60] = {};
        rep(i, n)
        {
            if (i == 0)
            {
                A[0][0] = b;
                A[0][1] = c;
            }
            else if (i == n - 1)
            {
                A[n - 1][n - 2] = a;
                A[n - 1][n - 1] = b;
            }
            else
            {
                A[i][i - 1] = a;
                A[i][i] = b;
                A[i][i + 1] = c;
            }
        }
        rep(i, n) AA[i][i] = 1;
        for (ll I = 0; 1LL << I <= t; I++)
        {
            if (t & (1 << I))
            {
                ll tmp[60][60] = {};
                rep(i, n) rep(j, n) rep(k, n)(tmp[i][j] += A[i][k] * AA[k][j]) %= m;
                rep(i, n) rep(j, n) AA[i][j] = tmp[i][j];
            }
            ll tmp[60][60] = {};
            rep(i, n) rep(j, n) rep(k, n)(tmp[i][j] += A[i][k] * A[k][j]) %= m;
            rep(i, n) rep(j, n) A[i][j] = tmp[i][j];
        }
        ll ans[60] = {};
        rep(i, n)
        {
            rep(j, n)
            {
                (ans[i] += AA[i][j] * S[j]) %= m;
            }
        }
        rep(i, n)
        {
            if (i == n - 1)
                cout << ans[i] << endl;
            else
                cout << ans[i] << " ";
        }
    }
}

### Prompt

In cpp, your task is to solve the following 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

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; i++)
using namespace std;
typedef long long ll;

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    while (1)
    {
        ll n, m, a, b, c, t;
        cin >> n >> m >> a >> b >> c >> t;
        if (n == 0)
            break;
        vector<ll> S(n);
        rep(i, n) cin >> S[i];
        ll A[60][60] = {}, AA[60][60] = {};
        rep(i, n)
        {
            if (i == 0)
            {
                A[0][0] = b;
                A[0][1] = c;
            }
            else if (i == n - 1)
            {
                A[n - 1][n - 2] = a;
                A[n - 1][n - 1] = b;
            }
            else
            {
                A[i][i - 1] = a;
                A[i][i] = b;
                A[i][i + 1] = c;
            }
        }
        rep(i, n) AA[i][i] = 1;
        for (ll I = 0; 1LL << I <= t; I++)
        {
            if (t & (1 << I))
            {
                ll tmp[60][60] = {};
                rep(i, n) rep(j, n) rep(k, n)(tmp[i][j] += A[i][k] * AA[k][j]) %= m;
                rep(i, n) rep(j, n) AA[i][j] = tmp[i][j];
            }
            ll tmp[60][60] = {};
            rep(i, n) rep(j, n) rep(k, n)(tmp[i][j] += A[i][k] * A[k][j]) %= m;
            rep(i, n) rep(j, n) A[i][j] = tmp[i][j];
        }
        ll ans[60] = {};
        rep(i, n)
        {
            rep(j, n)
            {
                (ans[i] += AA[i][j] * S[j]) %= m;
            }
        }
        rep(i, n)
        {
            if (i == n - 1)
                cout << ans[i] << endl;
            else
                cout << ans[i] << " ";
        }
    }
}

```