#include <cmath>
typedef long long ll;

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

#include <queue>
#include <cstdio>
inline int getInt(){ int s; scanf("%d", &s); return s; }

#include <set>
using namespace std;

typedef vector<ll> vec;
typedef vector<vec> mat;

int mod;

mat operator * (const mat &lhs, const mat &rhs){
  const int n = lhs.size();
  mat ret(n, vec(n));

  REP(k,n) REP(i,n) REP(j,n){
    ret[i][j] += (lhs[i][k] * rhs[k][j]) % mod;
    ret[i][j] %= mod;
  }

  return ret;
}

vec operator * (const mat &lhs, const vec &rhs){
  const int n = lhs.size();
  vec ret(n);

  REP(i,n) REP(j,n){
    ret[i] += (lhs[i][j] * rhs[j]) % mod;
    ret[i] %= mod;
  }

  return ret;
}

mat pow(mat m, int p){
  const int n = m.size();
  mat ret(n, vec(n));

  REP(i,n) ret[i][i] = 1;

  while(p){
    if(p & 1) ret = ret * m;
    m = m * m;
    p >>= 1;
  }

  return ret;
}

int main(){
  while(true){
    const int n = getInt();
    const int m = getInt();
    const int a = getInt();
    const int b = getInt();
    const int c = getInt();
    const int t = getInt();

    if(n + m + a + b + c + t == 0) break;

    mod = m;

    vec s(n);
    REP(i,n) s[i] = getInt();
    mat mt(n, vec(n));

    REP(i,n){
      if(i - 1 >= 0) mt[i][i - 1] = a;
      mt[i][i] = b;
      if(i + 1 < n) mt[i][i + 1] = c;
    }

    vec ans = pow(mt, t) * s;

    REP(i,n) printf("%d%c", (int)ans[i], i == n - 1 ? '\n' : ' ');
  }
  return 0;
}
### Prompt

Generate 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 <cmath>
typedef long long ll;

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

#include <queue>
#include <cstdio>
inline int getInt(){ int s; scanf("%d", &s); return s; }

#include <set>
using namespace std;

typedef vector<ll> vec;
typedef vector<vec> mat;

int mod;

mat operator * (const mat &lhs, const mat &rhs){
  const int n = lhs.size();
  mat ret(n, vec(n));

  REP(k,n) REP(i,n) REP(j,n){
    ret[i][j] += (lhs[i][k] * rhs[k][j]) % mod;
    ret[i][j] %= mod;
  }

  return ret;
}

vec operator * (const mat &lhs, const vec &rhs){
  const int n = lhs.size();
  vec ret(n);

  REP(i,n) REP(j,n){
    ret[i] += (lhs[i][j] * rhs[j]) % mod;
    ret[i] %= mod;
  }

  return ret;
}

mat pow(mat m, int p){
  const int n = m.size();
  mat ret(n, vec(n));

  REP(i,n) ret[i][i] = 1;

  while(p){
    if(p & 1) ret = ret * m;
    m = m * m;
    p >>= 1;
  }

  return ret;
}

int main(){
  while(true){
    const int n = getInt();
    const int m = getInt();
    const int a = getInt();
    const int b = getInt();
    const int c = getInt();
    const int t = getInt();

    if(n + m + a + b + c + t == 0) break;

    mod = m;

    vec s(n);
    REP(i,n) s[i] = getInt();
    mat mt(n, vec(n));

    REP(i,n){
      if(i - 1 >= 0) mt[i][i - 1] = a;
      mt[i][i] = b;
      if(i + 1 < n) mt[i][i + 1] = c;
    }

    vec ans = pow(mt, t) * s;

    REP(i,n) printf("%d%c", (int)ans[i], i == n - 1 ? '\n' : ' ');
  }
  return 0;
}
```