#include <bits/stdc++.h>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define REP(i, n) rep(i, 0, n)
#define repb(i, a, b) for(int i = a; i >= b; i--)
#define all(a) a.begin(), a.end()
#define int long long
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
using namespace std;
typedef pair<int, int> P;
const int mod = 1000000007;
const int INF = 1e12;

typedef vector<vector<int> > mat;
int M;

mat mul(mat &A, mat &B){
    mat C(A.size(), vector<int>(B[0].size()));
    rep(i, 0, A.size()){
        rep(k, 0, B.size()){
            rep(j, 0, B[0].size()){
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;
            }
        }
    }
    return C;
}
mat pow(mat A, int n){
    mat B(A.size(), vector<int>(A.size()));
    rep(i, 0, A.size()){
        B[i][i] = 1;
    }
    while(n > 0){
        if(n & 1) B = mul(B, A);
        A = mul(A, A);
        n >>= 1;
    }
    return B;
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m, a[3] = {}, t;
    while(1){
        cin >> n >> m;
        rep(i, 0, 3) cin >> a[i];
        cin >> t;
        M = m;
        if(n == 0 && m == 0 && a[0] == 0 && a[1] == 0 && a[2] == 0 && t == 0) break;
        vector<int> s(n);
        rep(i, 0, n) cin >> s[i];
        mat A(n, vector<int>(n));
        rep(i, 0, n){
            rep(j, 0, n){
                if(i - 1 == j) A[i][j] = a[0];
                if(i == j) A[i][j] = a[1];
                if(i + 1 == j) A[i][j] = a[2];
            }
        }
        A = pow(A, t);
        vector<int> ans(n);
        rep(i, 0, n){
            rep(j, 0, n){
                ans[i] = (ans[i] + A[i][j] * s[j]) % m; 
            }
        }
        for(int ii = 0; ii < ans.size(); ii++){
            if(ii) cout << ' ';
            cout << ans[ii];
        }
        cout << endl;
    }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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, a, n) for(int i = a; i < n; i++)
#define REP(i, n) rep(i, 0, n)
#define repb(i, a, b) for(int i = a; i >= b; i--)
#define all(a) a.begin(), a.end()
#define int long long
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
using namespace std;
typedef pair<int, int> P;
const int mod = 1000000007;
const int INF = 1e12;

typedef vector<vector<int> > mat;
int M;

mat mul(mat &A, mat &B){
    mat C(A.size(), vector<int>(B[0].size()));
    rep(i, 0, A.size()){
        rep(k, 0, B.size()){
            rep(j, 0, B[0].size()){
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;
            }
        }
    }
    return C;
}
mat pow(mat A, int n){
    mat B(A.size(), vector<int>(A.size()));
    rep(i, 0, A.size()){
        B[i][i] = 1;
    }
    while(n > 0){
        if(n & 1) B = mul(B, A);
        A = mul(A, A);
        n >>= 1;
    }
    return B;
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m, a[3] = {}, t;
    while(1){
        cin >> n >> m;
        rep(i, 0, 3) cin >> a[i];
        cin >> t;
        M = m;
        if(n == 0 && m == 0 && a[0] == 0 && a[1] == 0 && a[2] == 0 && t == 0) break;
        vector<int> s(n);
        rep(i, 0, n) cin >> s[i];
        mat A(n, vector<int>(n));
        rep(i, 0, n){
            rep(j, 0, n){
                if(i - 1 == j) A[i][j] = a[0];
                if(i == j) A[i][j] = a[1];
                if(i + 1 == j) A[i][j] = a[2];
            }
        }
        A = pow(A, t);
        vector<int> ans(n);
        rep(i, 0, n){
            rep(j, 0, n){
                ans[i] = (ans[i] + A[i][j] * s[j]) % m; 
            }
        }
        for(int ii = 0; ii < ans.size(); ii++){
            if(ii) cout << ' ';
            cout << ans[ii];
        }
        cout << endl;
    }
}

```