#include<bits/stdc++.h>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) for(int i = 0; i < (b); i++)
#define all(a) (a).begin(), (a).end()
#define show(x)  cerr << #x << " = " << (x) << endl;
using namespace std;

int M;
typedef vector<vector<int>> mat;

mat mul(mat &a, mat &b){
	mat c(a.size(), vector<int>(b[0].size()));
	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;
}

mat pow(mat a, int n){
	mat b(a.size(), vector<int>(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;
}

int n, x, b, c, t;

int solve(mat o){
	mat a(n, vector<int>(n));
	a[0][0] = b;
	a[0][1] = c;
	range(i,1,n - 1){
		a[i][i - 1] = x;
		a[i][i] = b;
		a[i][i + 1] = c;
	}
	a[n - 1][n - 2] = x;
	a[n - 1][n - 1] = b;

	//for(auto i : a){ for(auto j : i){ cout << j << ' '; } cout << endl; } cout << endl;

	a = pow(a,t); //??????A???n??????
	//for(auto i : a){ for(auto j : i){ cout << j << ' '; } cout << endl; } cout << endl;
	o = mul(a,o);
	rep(i,n){
		if(i) cout << ' ';
		cout << o[i][0];
	}
	cout << endl;
	//return a[1][0];
	return 1;
}

int main(){
	while(cin >> n >> M >> x >> b >> c >> t, n){
		mat o(n, vector<int>(1));
		rep(i,n){
			cin >> o[i][0];
		}
		solve(o);
	}
}
### Prompt

Develop a solution in Cpp to the problem described below:

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<bits/stdc++.h>
#define range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) for(int i = 0; i < (b); i++)
#define all(a) (a).begin(), (a).end()
#define show(x)  cerr << #x << " = " << (x) << endl;
using namespace std;

int M;
typedef vector<vector<int>> mat;

mat mul(mat &a, mat &b){
	mat c(a.size(), vector<int>(b[0].size()));
	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;
}

mat pow(mat a, int n){
	mat b(a.size(), vector<int>(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;
}

int n, x, b, c, t;

int solve(mat o){
	mat a(n, vector<int>(n));
	a[0][0] = b;
	a[0][1] = c;
	range(i,1,n - 1){
		a[i][i - 1] = x;
		a[i][i] = b;
		a[i][i + 1] = c;
	}
	a[n - 1][n - 2] = x;
	a[n - 1][n - 1] = b;

	//for(auto i : a){ for(auto j : i){ cout << j << ' '; } cout << endl; } cout << endl;

	a = pow(a,t); //??????A???n??????
	//for(auto i : a){ for(auto j : i){ cout << j << ' '; } cout << endl; } cout << endl;
	o = mul(a,o);
	rep(i,n){
		if(i) cout << ' ';
		cout << o[i][0];
	}
	cout << endl;
	//return a[1][0];
	return 1;
}

int main(){
	while(cin >> n >> M >> x >> b >> c >> t, n){
		mat o(n, vector<int>(1));
		rep(i,n){
			cin >> o[i][0];
		}
		solve(o);
	}
}
```