#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;
const double EPS = 1e-9;
typedef pair<int, int> P;
typedef unsigned int ui;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef long double ld;
const ld eps=1e-8;
int main() {
	int n, m, a, b, c, t;
	while (cin >> n >> m >> a >> b >> c >> t,n) {
		int out[52] = {};
		int ori[52] = {};
		rep1(i, n) {
			cin >> ori[i];
		}
		int tra[40][52][52] = {};
		rep1(i, n) {
			tra[0][i][i - 1] = c;
			tra[0][i][i] = b;
			tra[0][i][i + 1] = a;
		}
		rep(i, 32) {
			rep1(j, n) {
				rep1(k, n) {
					rep1(l, n) {
						tra[i + 1][j][l] += tra[i][j][k] * tra[i][k][l]%m;
					}
				}
			}
			rep1(j, n) {
				rep1(k, n) {
					tra[i + 1][j][k] %= m;
				}
			}
		}
		rep1(i, n) {
			out[i] = ori[i];
		}
		rep(i, 32) {
			if (t&(1 << i)) {
				int z[52] = {};
				rep1(j, n) {
					rep1(k, n) {
						z[k] += out[j]*tra[i][j][k] % m;
					}
				}
				rep1(j, n) {
					out[j] = z[j]%m;
				}
			}
		}
		rep1(i, n) {
			cout << out[i];
			if (i <= n - 1) {
				cout << " ";
			}
		}
		cout << endl;
	}
	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<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;
const double EPS = 1e-9;
typedef pair<int, int> P;
typedef unsigned int ui;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef long double ld;
const ld eps=1e-8;
int main() {
	int n, m, a, b, c, t;
	while (cin >> n >> m >> a >> b >> c >> t,n) {
		int out[52] = {};
		int ori[52] = {};
		rep1(i, n) {
			cin >> ori[i];
		}
		int tra[40][52][52] = {};
		rep1(i, n) {
			tra[0][i][i - 1] = c;
			tra[0][i][i] = b;
			tra[0][i][i + 1] = a;
		}
		rep(i, 32) {
			rep1(j, n) {
				rep1(k, n) {
					rep1(l, n) {
						tra[i + 1][j][l] += tra[i][j][k] * tra[i][k][l]%m;
					}
				}
			}
			rep1(j, n) {
				rep1(k, n) {
					tra[i + 1][j][k] %= m;
				}
			}
		}
		rep1(i, n) {
			out[i] = ori[i];
		}
		rep(i, 32) {
			if (t&(1 << i)) {
				int z[52] = {};
				rep1(j, n) {
					rep1(k, n) {
						z[k] += out[j]*tra[i][j][k] % m;
					}
				}
				rep1(j, n) {
					out[j] = z[j]%m;
				}
			}
		}
		rep1(i, n) {
			cout << out[i];
			if (i <= n - 1) {
				cout << " ";
			}
		}
		cout << endl;
	}
	return 0;
}

```