#include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <sstream>
#include <map>
#include <set>

#define REP(i,k,n) for(int i=k;i<n;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1<<30
#define pb push_back
#define mp make_pair

using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int MOD;

struct Mat {
	vector<vector<ll> > dat;
	int n;

	Mat(int n) : n(n), dat(n, vector<ll>(n)) {}

	Mat(vector<vector<ll> > dat) : n(dat.size()), dat(dat) {}

	Mat I(int n) {
		Mat ret(n);
		rep(i, n) ret.dat[i][i] = 1;
		return ret;
	}

	Mat mul(Mat &b) {
        Mat ret(n);
		rep(i, n) rep(j, n) rep(k, n) (ret.dat[i][j] += dat[i][k] * b.dat[k][j]) %= MOD;
        return ret;
	}

	Mat pow(ll b) {
		Mat ret = I(n);
        for (Mat A = *this; b > 0; A = A.mul(A) , b /= 2) if (b & 1) ret = A.mul(ret);
        return ret;
	}
};

int main() {
	int n, m, a, b, c, t;
	while(cin >> n >> m >> a >> b >> c >> t) {
		if(n == 0 && m == 0 && a == 0 && b == 0 && c == 0 && t == 0) break;

		MOD = m;

		vector<int> s(n);
		rep(i, n) cin >> s[i];

		int s1 = -1, s2 = 0, s3 = 1;
		Mat mat(n);
		rep(i, n) {
			if(s1 >= 0) {
				mat.dat[i][s1] = a;
			}
			mat.dat[i][s2] = b;
			if(s3 < n) {
				mat.dat[i][s3] = c;
			}
			
			s1++; s2++; s3++;
		}

		Mat ret = mat.pow(t);

		rep(i, n) {
			ll sum = 0;
			rep(j, n) {
				sum += s[j] * ret.dat[i][j];
				sum %= MOD;
			}

			cout << sum;

			if(i == n-1) cout << endl;
			else cout << " ";
		}
	}

	return 0;
}
### Prompt

Your task is to create 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 <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <sstream>
#include <map>
#include <set>

#define REP(i,k,n) for(int i=k;i<n;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1<<30
#define pb push_back
#define mp make_pair

using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int MOD;

struct Mat {
	vector<vector<ll> > dat;
	int n;

	Mat(int n) : n(n), dat(n, vector<ll>(n)) {}

	Mat(vector<vector<ll> > dat) : n(dat.size()), dat(dat) {}

	Mat I(int n) {
		Mat ret(n);
		rep(i, n) ret.dat[i][i] = 1;
		return ret;
	}

	Mat mul(Mat &b) {
        Mat ret(n);
		rep(i, n) rep(j, n) rep(k, n) (ret.dat[i][j] += dat[i][k] * b.dat[k][j]) %= MOD;
        return ret;
	}

	Mat pow(ll b) {
		Mat ret = I(n);
        for (Mat A = *this; b > 0; A = A.mul(A) , b /= 2) if (b & 1) ret = A.mul(ret);
        return ret;
	}
};

int main() {
	int n, m, a, b, c, t;
	while(cin >> n >> m >> a >> b >> c >> t) {
		if(n == 0 && m == 0 && a == 0 && b == 0 && c == 0 && t == 0) break;

		MOD = m;

		vector<int> s(n);
		rep(i, n) cin >> s[i];

		int s1 = -1, s2 = 0, s3 = 1;
		Mat mat(n);
		rep(i, n) {
			if(s1 >= 0) {
				mat.dat[i][s1] = a;
			}
			mat.dat[i][s2] = b;
			if(s3 < n) {
				mat.dat[i][s3] = c;
			}
			
			s1++; s2++; s3++;
		}

		Mat ret = mat.pow(t);

		rep(i, n) {
			ll sum = 0;
			rep(j, n) {
				sum += s[j] * ret.dat[i][j];
				sum %= MOD;
			}

			cout << sum;

			if(i == n-1) cout << endl;
			else cout << " ";
		}
	}

	return 0;
}
```