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

Solution (in CPP):

Solution:
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <algorithm>
#include <utility>
#include <functional>
#include <sstream>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <climits>
#include <fstream>
#include <list>
using namespace std;
inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toStr(T x) { ostringstream sout; sout << x; return sout.str(); }
typedef vector<int> vi;
typedef vector<vi>  vvi;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define REP(i,n) FOR(i,0,(n)-1)
const double EPS = 1e-10;
const double PI = acos(-1.0);
const int INF = INT_MAX / 10;

vvi mul(vvi A, vvi B, ll M) {
	vvi C(A.size(), vi(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;
}

vvi pow(vvi A, ll n, ll M) {
	vvi B(A.size(), vi(A.size()));
	REP(i, A.size()) {
		B[i][i] = 1;
	}
	while (n > 0) {
		if (n & 1) {
			B = mul(B, A, M);
		}
		A = mul(A, A, M);
		n >>= 1;
	}
	return B;
}

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

		vvi S(N, vi(1));
		REP(i, N) {
			cin >> S[i][0];
		}

		vvi ans = mul(pow(mat, T, M), S, M);
		REP(i, N) {
			cout << ans[i][0];
			if (i != N - 1) {
				cout << " ";
			}
		}
		cout << endl;
	}

	return 0;
}