#include <iostream>
#include <complex>
#include <sstream>
#include <string>
#include <algorithm>
#include <deque>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <vector>
#include <set>
#include <limits>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <ctime>
using namespace std;
#define REP(i, j) for(int i = 0; i < (int)(j); ++i)
#define FOR(i, j, k) for(int i = (int)(j); i < (int)(k); ++i)
#define SORT(v) sort((v).begin(), (v).end())
#define REVERSE(v) reverse((v).begin(), (v).end())
typedef complex<double> P;
typedef vector<long long> vec;
typedef vector<vec> mat;

long long n;
long long H, W, M;

//A*B
mat mul(mat &A, mat &B){
  mat C(A.size(), vec(B[0].size(), 0));
  REP(i, A.size())
    REP(k, B.size())
      REP(j, B[0].size())
        C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;
  return C;
}

//A^n
mat pow(mat A, long long n){
  mat B(A.size(), vec(A.size()));
  REP(i, A.size())
    B[i][i] = 1;
  while(n > 0){
    if(n & 1) B = mul(B, A);
    A = mul(A, A);
    n >>= 1;
  }
  return B;
}

void solve(long long a, long long b, long long c, vector<long long> &S){
  mat A(H, vec(W));
  FOR(i, 1, H) A[i][i - 1] = a;
  REP(i, H) A[i][i] = b;
  FOR(i, 1, H) A[i - 1][i] = c;
  A = pow(A, n);
  REP(i, S.size()){
    int tmp = 0;
    REP(j, W) tmp = (tmp + S[j] * A[i][j]) % M;
    if(i != 0) cout <<" ";
    cout <<tmp;
  }
  cout <<endl;
}

int main() {
  long long N, A, B, C, T;
  while(cin >>N >>M >>A >>B >>C >>T && N){
    vector<long long> S(N);
    REP(i, N) cin >>S[i];
    H = W = N;
    n = T;
    solve(A, B, C, S);
  }
  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 <iostream>
#include <complex>
#include <sstream>
#include <string>
#include <algorithm>
#include <deque>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <vector>
#include <set>
#include <limits>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <ctime>
using namespace std;
#define REP(i, j) for(int i = 0; i < (int)(j); ++i)
#define FOR(i, j, k) for(int i = (int)(j); i < (int)(k); ++i)
#define SORT(v) sort((v).begin(), (v).end())
#define REVERSE(v) reverse((v).begin(), (v).end())
typedef complex<double> P;
typedef vector<long long> vec;
typedef vector<vec> mat;

long long n;
long long H, W, M;

//A*B
mat mul(mat &A, mat &B){
  mat C(A.size(), vec(B[0].size(), 0));
  REP(i, A.size())
    REP(k, B.size())
      REP(j, B[0].size())
        C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;
  return C;
}

//A^n
mat pow(mat A, long long n){
  mat B(A.size(), vec(A.size()));
  REP(i, A.size())
    B[i][i] = 1;
  while(n > 0){
    if(n & 1) B = mul(B, A);
    A = mul(A, A);
    n >>= 1;
  }
  return B;
}

void solve(long long a, long long b, long long c, vector<long long> &S){
  mat A(H, vec(W));
  FOR(i, 1, H) A[i][i - 1] = a;
  REP(i, H) A[i][i] = b;
  FOR(i, 1, H) A[i - 1][i] = c;
  A = pow(A, n);
  REP(i, S.size()){
    int tmp = 0;
    REP(j, W) tmp = (tmp + S[j] * A[i][j]) % M;
    if(i != 0) cout <<" ";
    cout <<tmp;
  }
  cout <<endl;
}

int main() {
  long long N, A, B, C, T;
  while(cin >>N >>M >>A >>B >>C >>T && N){
    vector<long long> S(N);
    REP(i, N) cin >>S[i];
    H = W = N;
    n = T;
    solve(A, B, C, S);
  }
  return 0;
}
```