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

Solution (in CPP):

Solution:
class in{struct It{int it;const bool rev;explicit constexpr It(int it_, bool rev=false):it(it_),rev(rev){}int operator*(){return it;}bool operator!=(It& r){return it!=r.it;}void operator++(){rev?--it:++it;}};const It i,n;public:explicit constexpr in(int n):i(0),n(n<0?0:n){}explicit constexpr in(int i,int n):i(i,n<i),n(n){}const It& begin(){return i;}const It& end(){return n;}};
  
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

typedef long long i64;
typedef vector<i64> vec;
typedef vector<vec> mat;
i64 MOD;
//A * B
mat mat_mul(const mat& A, const mat& B) {
    mat C(A.size(), vec(B.size()));
    for(size_t i = 0; i < A.size(); ++i)
      for(size_t k = 0; k < B.size(); ++k)
        for(size_t j = 0; j < B[0].size(); ++j)
          C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
    return C;
}
//A ^ n
mat mat_pow(mat A, i64 n, i64 mod) {
    MOD = mod;
    mat B(A.size(), vec(A.size(), 0LL));
    for(size_t i = 0; i < A.size(); ++i) B[i][i] = 1LL;
    while(n > 0) {
        if(n & 1LL) B = mat_mul(B, A);
        A = mat_mul(A, A);
        n >>= 1LL;
    }
    return B;
}

int main() {
    int n, mod, a, b, c, t;
    while(cin >> n >> mod >> a >> b >> c >> t && n) {
        mat B(n, vec(n, 0));
        vec A(n, 0);
        
        for(auto& x : A) cin >> x;
        for(int i : in(n)) {
            if(i > 0)
                B[i][i - 1] = a;
            B[i][i] = b;
            if(i < n - 1)
                B[i][i + 1] = c;
        }
        B = mat_pow(B, t, mod);
        for(int i : in(n)) {
            i64 ans = 0LL;
            for(int j : in(n)) (ans += B[i][j] * A[j]) %= mod;
            cout << ans << (i < n - 1 ? ' ' : '\n');
        }
    }
    return 0;
}