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

typedef long long ll;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define each(itr,c) for(__typeof(c.begin()) itr=c.begin(); itr!=c.end(); ++itr)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second

typedef vector<int> vi;
typedef vector<vi> Mat;

Mat mul(const Mat &a, const Mat &b, int mod)
{
    int n=a.size();
    Mat c(n,vi(n,0));
    rep(i,n)rep(j,n)rep(k,n) (c[i][j]+=a[i][k]*b[k][j])%=mod;
    return c;
}

Mat make(int n, int A, int B, int C)
{
    Mat r(n,vi(n,0));
    if(n==1) r[0][0]=B;
    else
    {
        r[0][0]=B;
        r[0][1]=C;
        r[n-1][n-2]=A;
        r[n-1][n-1]=B;
        for(int i=1; i<n-1; ++i)
        {
            r[i][i-1]=A;
            r[i][i]=B;
            r[i][i+1]=C;
        }
    }
    return r;
}

int main()
{
    int n,m,a,b,c,t;
    while(cin>>n>>m>>a>>b>>c>>t,n)
    {
        vector<int> s(n);
        rep(i,n) cin >>s[i];

        Mat P[31];
        P[0]=make(n,a,b,c);
        for(int i=1; i<31; ++i) P[i]=mul(P[i-1],P[i-1],m);

        Mat A(n,vi(n,0));
        rep(i,n) A[i][i]=1;
        for(int i=30; i>=0; --i)
        {
            if(t>=(1<<i))
            {
                A=mul(A,P[i],m);
                t-=(1<<i);
            }
        }

        rep(i,n)
        {
            int ans=0;
            rep(j,n) (ans+=A[i][j]*s[j])%=m;
            if(i) printf(" ");
            printf("%d", ans);
        }
        printf("\n");
    }
    return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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;

typedef long long ll;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define each(itr,c) for(__typeof(c.begin()) itr=c.begin(); itr!=c.end(); ++itr)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second

typedef vector<int> vi;
typedef vector<vi> Mat;

Mat mul(const Mat &a, const Mat &b, int mod)
{
    int n=a.size();
    Mat c(n,vi(n,0));
    rep(i,n)rep(j,n)rep(k,n) (c[i][j]+=a[i][k]*b[k][j])%=mod;
    return c;
}

Mat make(int n, int A, int B, int C)
{
    Mat r(n,vi(n,0));
    if(n==1) r[0][0]=B;
    else
    {
        r[0][0]=B;
        r[0][1]=C;
        r[n-1][n-2]=A;
        r[n-1][n-1]=B;
        for(int i=1; i<n-1; ++i)
        {
            r[i][i-1]=A;
            r[i][i]=B;
            r[i][i+1]=C;
        }
    }
    return r;
}

int main()
{
    int n,m,a,b,c,t;
    while(cin>>n>>m>>a>>b>>c>>t,n)
    {
        vector<int> s(n);
        rep(i,n) cin >>s[i];

        Mat P[31];
        P[0]=make(n,a,b,c);
        for(int i=1; i<31; ++i) P[i]=mul(P[i-1],P[i-1],m);

        Mat A(n,vi(n,0));
        rep(i,n) A[i][i]=1;
        for(int i=30; i>=0; --i)
        {
            if(t>=(1<<i))
            {
                A=mul(A,P[i],m);
                t-=(1<<i);
            }
        }

        rep(i,n)
        {
            int ans=0;
            rep(j,n) (ans+=A[i][j]*s[j])%=m;
            if(i) printf(" ");
            printf("%d", ans);
        }
        printf("\n");
    }
    return 0;
}
```