Problem:
Problem C: One-Dimensional Cellular Automaton

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.

Sample 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 for the Sample Input

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 C++ (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <vector>
#include <complex>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <stack>
#include <cmath>
#include <iomanip>
#include <sstream>

using namespace std;

typedef long long ll;
typedef ll li;
typedef pair<int,int> PI;

#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define REP(i,m,n) for(int i=m; i<=(int)(n); ++i)
#define F first
#define S second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define SZ(a) (int)((a).size())
#define ALL(a) a.begin(),a.end()
#define FOR(it,a) for(__typeof(a.begin())it=a.begin();it!=a.end();++it)
int dx[]={0,1,0,-1,1,1,-1,-1};
int dy[]={1,0,-1,0,-1,1,1,-1};


int n,m,a,b,c,t;
int S[50];

typedef vector<vector<int> > mat;

mat matmul(mat a,mat b){
  mat c(n,vector<int>(n,0));
  rep(i,n) rep(j,n) rep(k,n)
    c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % m;
  return c;
}

mat matpow(mat a,int ex){
  mat ret(n,vector<int>(n,0));
  rep(i,n) ret[i][i] = 1;
  mat base(a);

  while(ex){
    if(ex&1) ret = matmul(ret,base);
    base=matmul(base,base);
    ex>>=1;
  }
  return ret;
}

void solve(){
  rep(i,n) cin >> S[i];
  mat abc(n,vector<int>(n));
  rep(i,n){
    if(i) abc[i][i-1]=a;
    abc[i][i]=b;
    if(i+1<n) abc[i][i+1]=c;
  }

  abc = matpow(abc,t);

  vector<int> ans(n,0);
  
  rep(i,n)rep(j,n)
    ans[i] = (ans[i] + abc[i][j] * S[j]) % m;
  rep(i,n){
    if(i) cout << ' ';
    cout << ans[i];
  }
  cout << endl;

}
  

main(){
  while(cin >> n >> m >> a >> b >> c >> t,n)
    solve();
}