#include <bits/stdc++.h>
using namespace std;

#define rep(i,a,b) for(int i=a;i<b;i++)

typedef long long int ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;

int M;

vi mul_mat_vec(vvi a, vi b)
{
	int n = b.size();
	vi ret(n, 0);
	rep(i, 0, n)
	{
		ret[i] = 0;
		rep(j, 0, n) ret[i] += a[i][j] * b[j];
		ret[i] %= M;
	}
	return ret;
}

vvi mul_mat_mat(vvi a, vvi b)
{
	int n = a.size();
	vvi ret(n, vi(n, 0));
	rep(i, 0, n) rep(j, 0, n)
	{
		ret[i][j] = 0;
		rep(k, 0, n) ret[i][j] += a[i][k] * b[k][j];
		ret[i][j] %= M;
	}
	return ret;
}

void write_vec(vi a)
{
	int n = a.size();
	rep(i, 0, n) cout << a[i] << " ";
	cout << endl;
}

void write_mat(vvi a)
{
	int n = a.size();
	int nn = a[0].size();
	rep(i, 0, n)
	{
		rep(j, 0, nn) cout << a[i][j] << " ";
		cout << endl;
	}
}

vvi fastpow(vvi x, int n)
{
	vvi ret(x.size(), vi(x.size(), 0));
	rep(i, 0, x.size()) ret[i][i] = 1;

	while (0 < n)
	{
		if ((n % 2) == 0)
		{
			x = mul_mat_mat(x, x);
			n >>= 1;
		}
		else
		{
			ret = mul_mat_mat(ret, x);
			--n;
		}
	}
	return ret;
}

int main()
{
	while (1)
	{
		int N, A, B, C, T;
		cin >> N >> M >> A >> B >> C >> T;
		if (N == 0) return 0;
		vi S(N);
		rep(i, 0, N) cin >> S[i];
		
		vvi mat(N, vi(N, 0));
		rep(i, 0, N)
		{
			if (i != 0) mat[i][i - 1] = A;
			mat[i][i] = B;
			if (i != N - 1) mat[i][i + 1] = C;
		}


		
		vvi mats = fastpow(mat, T);
		S = mul_mat_vec(mats, S);

		cout << S[0];
		rep(i, 1, N) cout << " " << S[i];
		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 <bits/stdc++.h>
using namespace std;

#define rep(i,a,b) for(int i=a;i<b;i++)

typedef long long int ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;

int M;

vi mul_mat_vec(vvi a, vi b)
{
	int n = b.size();
	vi ret(n, 0);
	rep(i, 0, n)
	{
		ret[i] = 0;
		rep(j, 0, n) ret[i] += a[i][j] * b[j];
		ret[i] %= M;
	}
	return ret;
}

vvi mul_mat_mat(vvi a, vvi b)
{
	int n = a.size();
	vvi ret(n, vi(n, 0));
	rep(i, 0, n) rep(j, 0, n)
	{
		ret[i][j] = 0;
		rep(k, 0, n) ret[i][j] += a[i][k] * b[k][j];
		ret[i][j] %= M;
	}
	return ret;
}

void write_vec(vi a)
{
	int n = a.size();
	rep(i, 0, n) cout << a[i] << " ";
	cout << endl;
}

void write_mat(vvi a)
{
	int n = a.size();
	int nn = a[0].size();
	rep(i, 0, n)
	{
		rep(j, 0, nn) cout << a[i][j] << " ";
		cout << endl;
	}
}

vvi fastpow(vvi x, int n)
{
	vvi ret(x.size(), vi(x.size(), 0));
	rep(i, 0, x.size()) ret[i][i] = 1;

	while (0 < n)
	{
		if ((n % 2) == 0)
		{
			x = mul_mat_mat(x, x);
			n >>= 1;
		}
		else
		{
			ret = mul_mat_mat(ret, x);
			--n;
		}
	}
	return ret;
}

int main()
{
	while (1)
	{
		int N, A, B, C, T;
		cin >> N >> M >> A >> B >> C >> T;
		if (N == 0) return 0;
		vi S(N);
		rep(i, 0, N) cin >> S[i];
		
		vvi mat(N, vi(N, 0));
		rep(i, 0, N)
		{
			if (i != 0) mat[i][i - 1] = A;
			mat[i][i] = B;
			if (i != N - 1) mat[i][i + 1] = C;
		}


		
		vvi mats = fastpow(mat, T);
		S = mul_mat_vec(mats, S);

		cout << S[0];
		rep(i, 1, N) cout << " " << S[i];
		cout << endl;
	}
}
```