#include <bits/stdc++.h>
#define For(i, a, b) for(int (i)=(a); (i)<(b); ++(i))
#define rFor(i, a, b) for(int (i)=(a)-1; (i)>=(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef pair<int, int> pii;
 
typedef vector<vector<lint>> mat;
 
mat mul(mat &a, mat &b, lint mod){
    int n = a.size();
    int m = b[0].size();
    mat c = vector<vector<lint>>(n, vector<lint>(m, 0));
    rep(i, n)rep(j, m){
        lint t = 0;
        rep(k, b.size()) t = (t + a[i][k] * b[k][j] % mod) % mod;
        c[i][j] = t;
    }
    return c;
}
 
mat pow(mat a, lint k, lint mod){
    int n = a.size();
    mat b = vector<vector<lint>>(n, vector<lint>(n, 0));
    rep(i, n) b[i][i] = 1;
    while(k){
        if(k & 1) b = mul(b, a, mod), k -= 1;
        else a = mul(a, a, mod), k >>= 1;
    }
    return b;
}

lint powmod(lint x, lint n, lint mod){
  lint ret = 1;
  while(n > 0){
    if(n&1) ret *= x, ret %= mod, n -= 1;
    else x *= x, x %= mod, n >>= 1;
  }
  return ret;
}

int main(){
    int n, m, a, b, c, t;
    while(scanf("%d", &n) && n){
        scanf("%d%d%d%d%d", &m, &a, &b, &c, &t);
        if(n == 1){
            int s;
            scanf("%d", &s);
            printf("%d\n", s * powmod(b, t, m) % m);
            continue;
        }
        mat A = vector<vector<lint>>(n, vector<lint>(n, 0));
        mat s = vector<vector<lint>>(n, vector<lint>(1, 0));
        rep(i, n) scanf("%d", &s[i][0]);
        A[0][0] = A[n-1][n-1] = b;
        A[0][1] = c;
        A[n-1][n-2] = a;
        For(i, 1, n-1){
            A[i][i-1] = a;
            A[i][i] = b;
            A[i][i+1] = c;
        }
        A = pow(A, t, m);
        s = mul(A, s, m);
        printf("%lld", s[0][0]);
        rep(i, n-1) printf(" %lld", s[i+1][0]);
        printf("\n");
    }
}

### Prompt

Please create a solution in CPP to 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 For(i, a, b) for(int (i)=(a); (i)<(b); ++(i))
#define rFor(i, a, b) for(int (i)=(a)-1; (i)>=(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef pair<int, int> pii;
 
typedef vector<vector<lint>> mat;
 
mat mul(mat &a, mat &b, lint mod){
    int n = a.size();
    int m = b[0].size();
    mat c = vector<vector<lint>>(n, vector<lint>(m, 0));
    rep(i, n)rep(j, m){
        lint t = 0;
        rep(k, b.size()) t = (t + a[i][k] * b[k][j] % mod) % mod;
        c[i][j] = t;
    }
    return c;
}
 
mat pow(mat a, lint k, lint mod){
    int n = a.size();
    mat b = vector<vector<lint>>(n, vector<lint>(n, 0));
    rep(i, n) b[i][i] = 1;
    while(k){
        if(k & 1) b = mul(b, a, mod), k -= 1;
        else a = mul(a, a, mod), k >>= 1;
    }
    return b;
}

lint powmod(lint x, lint n, lint mod){
  lint ret = 1;
  while(n > 0){
    if(n&1) ret *= x, ret %= mod, n -= 1;
    else x *= x, x %= mod, n >>= 1;
  }
  return ret;
}

int main(){
    int n, m, a, b, c, t;
    while(scanf("%d", &n) && n){
        scanf("%d%d%d%d%d", &m, &a, &b, &c, &t);
        if(n == 1){
            int s;
            scanf("%d", &s);
            printf("%d\n", s * powmod(b, t, m) % m);
            continue;
        }
        mat A = vector<vector<lint>>(n, vector<lint>(n, 0));
        mat s = vector<vector<lint>>(n, vector<lint>(1, 0));
        rep(i, n) scanf("%d", &s[i][0]);
        A[0][0] = A[n-1][n-1] = b;
        A[0][1] = c;
        A[n-1][n-2] = a;
        For(i, 1, n-1){
            A[i][i-1] = a;
            A[i][i] = b;
            A[i][i+1] = c;
        }
        A = pow(A, t, m);
        s = mul(A, s, m);
        printf("%lld", s[0][0]);
        rep(i, n-1) printf(" %lld", s[i+1][0]);
        printf("\n");
    }
}

```