#include <stdio.h>
#include <cctype>
#include <limits.h>
#include <math.h>
#include <complex>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cstring>
#include <string>
#include <sstream>
#include <algorithm>
#include <iomanip>
#include <iostream>

#define VARIABLE(x) cerr << #x << "=" << x << endl
#define BINARY(x) static_cast<bitset<16> >(x);
#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 if_range(x, y, w, h) if (0<=(int)(x) && (int)(x)<(int)(w) && 0<=(int)(y) && (int)(y)<(int)(h))

const int INF = 1000000000;
const double EPS = 1e-8;
const double PI = 3.14159;
int dx[4]={0, 1, 0, -1}, dy[4]={-1, 0, 1, 0};
using namespace std;
typedef long long ll;
//typedef pair<int, int> P;
struct P {
	int x, y;
	P(int x, int y):x(x), y(y){}
	P(){}
};

typedef vector<int> Vec;
typedef vector<Vec> Mat;

int N, mod;

Mat mul(Mat &A, Mat &B)
{
	Mat C(A.size(), Vec(B[0].size()));
	for (int i=0; i<A.size(); i++) {
		for (int k=0; k<B.size(); k++) {
			for (int j=0; j<B[0].size(); j++) {
				C[i][j] += A[i][k] * B[k][j];
				C[i][j] %= mod;
			}
		}
	}
	return C;
}

Mat pow(Mat A, ll y)
{
	Mat B(A.size(), Vec(A.size()));
	for (int i=0; i<A.size(); i++) B[i][i]=1;
	
	for (; y>0; y>>=1, A=mul(A, A)) {
		if (y&1) B = mul(B, A);
	}
	
	return B;
}


/** Problem1327 : One-Dimensional Cellular Automaton **/
int main()
{
	int A, B, C; ll T;
	while (cin>>N>>mod>>A>>B>>C>>T, N||mod||A||B||C||T) {
		Mat m(N, Vec(N));
		Vec S(N);
		
		rep(i, N) cin>>S[i];
		
		int base=0;
		rep(i, N) {
			rep(j, N) {
				if (j==base-1) m[i][j] = A;
				else if (j==base) m[i][j] = B;
				else if (j==base+1) m[i][j] = C;
				else m[i][j] = 0;
			}
			base++;
		}
		
		m = pow(m, T);
		
		rep(i, N) {
			int t=0;
			rep(j, N) {
				t += S[j] * m[i][j];
				t %= mod;
			}
			if (i!=0) cout<<" ";
			cout << t;
		}
		cout << endl;
	}
}
### 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 <stdio.h>
#include <cctype>
#include <limits.h>
#include <math.h>
#include <complex>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cstring>
#include <string>
#include <sstream>
#include <algorithm>
#include <iomanip>
#include <iostream>

#define VARIABLE(x) cerr << #x << "=" << x << endl
#define BINARY(x) static_cast<bitset<16> >(x);
#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 if_range(x, y, w, h) if (0<=(int)(x) && (int)(x)<(int)(w) && 0<=(int)(y) && (int)(y)<(int)(h))

const int INF = 1000000000;
const double EPS = 1e-8;
const double PI = 3.14159;
int dx[4]={0, 1, 0, -1}, dy[4]={-1, 0, 1, 0};
using namespace std;
typedef long long ll;
//typedef pair<int, int> P;
struct P {
	int x, y;
	P(int x, int y):x(x), y(y){}
	P(){}
};

typedef vector<int> Vec;
typedef vector<Vec> Mat;

int N, mod;

Mat mul(Mat &A, Mat &B)
{
	Mat C(A.size(), Vec(B[0].size()));
	for (int i=0; i<A.size(); i++) {
		for (int k=0; k<B.size(); k++) {
			for (int j=0; j<B[0].size(); j++) {
				C[i][j] += A[i][k] * B[k][j];
				C[i][j] %= mod;
			}
		}
	}
	return C;
}

Mat pow(Mat A, ll y)
{
	Mat B(A.size(), Vec(A.size()));
	for (int i=0; i<A.size(); i++) B[i][i]=1;
	
	for (; y>0; y>>=1, A=mul(A, A)) {
		if (y&1) B = mul(B, A);
	}
	
	return B;
}


/** Problem1327 : One-Dimensional Cellular Automaton **/
int main()
{
	int A, B, C; ll T;
	while (cin>>N>>mod>>A>>B>>C>>T, N||mod||A||B||C||T) {
		Mat m(N, Vec(N));
		Vec S(N);
		
		rep(i, N) cin>>S[i];
		
		int base=0;
		rep(i, N) {
			rep(j, N) {
				if (j==base-1) m[i][j] = A;
				else if (j==base) m[i][j] = B;
				else if (j==base+1) m[i][j] = C;
				else m[i][j] = 0;
			}
			base++;
		}
		
		m = pow(m, T);
		
		rep(i, N) {
			int t=0;
			rep(j, N) {
				t += S[j] * m[i][j];
				t %= mod;
			}
			if (i!=0) cout<<" ";
			cout << t;
		}
		cout << endl;
	}
}
```