#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <sstream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <numeric>
#include <cctype>
#include <cassert>

// BEGIN CUT HERE
#ifdef _MSC_VER
#include <agents.h>
#endif
// END CUT HERE 

#define FOR(i, a, b) for(int i = (a); i < (int)(b); ++i)
#define rep(i, n) FOR(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define REV(v) v.rbegin(), v.rend()
#define MEMSET(v, s) memset(v, s, sizeof(v))
#define MP make_pair
#define MT make_tuple
#define X first
#define Y second

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;

int n, m, t;
int a[3];
vector<vector<int>> mul(vector<vector<int>> &a, vector<vector<int>> &b){
	int r1 = a.size(), c1 = a[0].size();
	int r2 = b.size(), c2 = b[0].size();
	assert(c1 == r2);
	auto res = vector<vector<int>>(r1, vector<int>(c2));

	rep(i, r1) rep(j, c2){
		int sum = 0;
		rep(k, c1){
			sum = (sum + a[i][k] * b[k][j]) % m;
		}
		res[i][j] = sum;
	}

	return res;
}

vector<vector<int>> mod_pow(vector<vector<int>> v, int p){
	auto res = v;
	rep(i, n) rep(j, n) res[i][j] = i == j;

	while (p){
		if (p & 1) res = mul(res, v);
		v = mul(v, v);
		p /= 2;
	}

	return res;
}

int main(){
	while (cin >> n >> m >> a[0] >> a[1] >> a[2] >> t, n){
		vector<vector<int>> ss(n);
		rep(i, n){
			int s;
			cin >> s;
			ss[i].push_back(s);
		}

		auto v = vector<vector<int>>(n, vector<int>(n, 0));
		rep(i, n){
			for (int j = -1; j <= 1; ++j){
				if (0 <= i + j && i + j < n) v[i][i + j] = a[j + 1];
			}
		}

		auto tmp = mod_pow(v, t);
		auto ans = mul(tmp, ss);

		bool f = true;
		for (auto v : ans) for (auto e : v){
			if (!f) cout << " ";
			f = false;
			cout << e;
		}
		cout << endl;
	}

	return 0;
}
### 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 <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <sstream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <numeric>
#include <cctype>
#include <cassert>

// BEGIN CUT HERE
#ifdef _MSC_VER
#include <agents.h>
#endif
// END CUT HERE 

#define FOR(i, a, b) for(int i = (a); i < (int)(b); ++i)
#define rep(i, n) FOR(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define REV(v) v.rbegin(), v.rend()
#define MEMSET(v, s) memset(v, s, sizeof(v))
#define MP make_pair
#define MT make_tuple
#define X first
#define Y second

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;

int n, m, t;
int a[3];
vector<vector<int>> mul(vector<vector<int>> &a, vector<vector<int>> &b){
	int r1 = a.size(), c1 = a[0].size();
	int r2 = b.size(), c2 = b[0].size();
	assert(c1 == r2);
	auto res = vector<vector<int>>(r1, vector<int>(c2));

	rep(i, r1) rep(j, c2){
		int sum = 0;
		rep(k, c1){
			sum = (sum + a[i][k] * b[k][j]) % m;
		}
		res[i][j] = sum;
	}

	return res;
}

vector<vector<int>> mod_pow(vector<vector<int>> v, int p){
	auto res = v;
	rep(i, n) rep(j, n) res[i][j] = i == j;

	while (p){
		if (p & 1) res = mul(res, v);
		v = mul(v, v);
		p /= 2;
	}

	return res;
}

int main(){
	while (cin >> n >> m >> a[0] >> a[1] >> a[2] >> t, n){
		vector<vector<int>> ss(n);
		rep(i, n){
			int s;
			cin >> s;
			ss[i].push_back(s);
		}

		auto v = vector<vector<int>>(n, vector<int>(n, 0));
		rep(i, n){
			for (int j = -1; j <= 1; ++j){
				if (0 <= i + j && i + j < n) v[i][i + j] = a[j + 1];
			}
		}

		auto tmp = mod_pow(v, t);
		auto ans = mul(tmp, ss);

		bool f = true;
		for (auto v : ans) for (auto e : v){
			if (!f) cout << " ";
			f = false;
			cout << e;
		}
		cout << endl;
	}

	return 0;
}
```