#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <algorithm>

#define REP(i,n) for(int i=0;i<(int)(n);i++)

using namespace std;
typedef int64_t Data;
typedef vector<Data> Array;
typedef vector<Array> Matrix;

int64_t MOD;

Matrix operator*(const Matrix &lhs, const Matrix &rhs) {
  Matrix res(lhs.size(), Array(rhs[0].size(), 0));
  REP(i,lhs.size()) REP(j,rhs[0].size()) REP(k,rhs.size()){
    res[i][j] = res[i][j] + lhs[i][k] * rhs[k][j];
    res[i][j] %= MOD;
  }
  return res;
}

Matrix scalar(int size, Data k) {
  Matrix mat(size, Array(size, 0));
  REP(i,size) mat[i][i] = k;
  return mat;
}

Matrix operator^(const Matrix &lhs, const int n) {
  if (n == 0) return scalar(lhs.size(), 1);
  Matrix res = (lhs * lhs) ^ (n / 2);
  if (n % 2) res = res * lhs;
  return res;
}

int main() {
  while(1){
    int n,m,a,b,c,t;
    cin>>n>>m>>a>>b>>c>>t;
    if(!n)break;
    MOD = m;
    vector<int> v(n);
    REP(i,n)cin>>v[i];
    Matrix mat(n,Array(n));
    REP(i,n){
      if(i)mat[i][i-1] = a;
      if(i)mat[i-1][i] = c;
      mat[i][i] = b;
    }
    auto res = mat ^ t;
    REP(i,n) {
      int64_t val = 0;
      REP(j,n){
        val += res[i][j] * v[j];
        val %= MOD;
      }
      if(i)cout<<' ';
      cout << val;
    }
    cout<<endl;
  }
  return 0;
}
### Prompt

Your task is to create a Cpp solution 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 <iostream>
#include <vector>
#include <string>
#include <map>
#include <algorithm>

#define REP(i,n) for(int i=0;i<(int)(n);i++)

using namespace std;
typedef int64_t Data;
typedef vector<Data> Array;
typedef vector<Array> Matrix;

int64_t MOD;

Matrix operator*(const Matrix &lhs, const Matrix &rhs) {
  Matrix res(lhs.size(), Array(rhs[0].size(), 0));
  REP(i,lhs.size()) REP(j,rhs[0].size()) REP(k,rhs.size()){
    res[i][j] = res[i][j] + lhs[i][k] * rhs[k][j];
    res[i][j] %= MOD;
  }
  return res;
}

Matrix scalar(int size, Data k) {
  Matrix mat(size, Array(size, 0));
  REP(i,size) mat[i][i] = k;
  return mat;
}

Matrix operator^(const Matrix &lhs, const int n) {
  if (n == 0) return scalar(lhs.size(), 1);
  Matrix res = (lhs * lhs) ^ (n / 2);
  if (n % 2) res = res * lhs;
  return res;
}

int main() {
  while(1){
    int n,m,a,b,c,t;
    cin>>n>>m>>a>>b>>c>>t;
    if(!n)break;
    MOD = m;
    vector<int> v(n);
    REP(i,n)cin>>v[i];
    Matrix mat(n,Array(n));
    REP(i,n){
      if(i)mat[i][i-1] = a;
      if(i)mat[i-1][i] = c;
      mat[i][i] = b;
    }
    auto res = mat ^ t;
    REP(i,n) {
      int64_t val = 0;
      REP(j,n){
        val += res[i][j] * v[j];
        val %= MOD;
      }
      if(i)cout<<' ';
      cout << val;
    }
    cout<<endl;
  }
  return 0;
}
```