Whenever rxw meets Coral, he requires her to give him the laboratory key. Coral does not want to give him the key, so Coral ask him one question. if rxw can solve the problem, she will give him the key, otherwise do not give him. rxw turns to you for help now,can you help him?N children form a circle, numbered 0,1,2, ... ..., n-1,with Clockwise. Initially the ith child has Ai apples. Each round game, the ith child will obtain ( L*A_(i+n-1)%n+R*A_(i+1)%n ) apples. After m rounds game, Coral would like to know the number of apples each child has. Because the final figure may be very large, so output the number model M.

The first line of input is an integer T representing the number of test cases to follow. Each case consists of two lines of input: the first line contains five integers n,m,L,R and M . the second line contains n integers A0, A1, A2 ... An-1. (0 <= Ai <= 1000,3 <= n <= 100,0 <= L, R <= 1000,1 <= M <= 10 ^ 6,0 <=m < = 10 ^ 9). After m rounds game, output the number model M of apples each child has.

Each case separated by a space. See sample.

1

3 2 3 4 10000

1 2 3

120 133 131

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<cstdlib>

 using namespace std;

 #define N 102

 typedef __int64 LL;

 LL n,m,L,R,mod;

 LL f[];

 struct Matrix

 {

     LL mat[N][N];

     void ini()

     {

         memset(mat,,sizeof(mat));

     }

     void init()

     {

         for(LL i=;i<n;i++)

         for(LL j=;j<n;j++)

         if(i==j) mat[i][j]=;

         else mat[i][j]=;

     }

     void make_first()

     {

         for(LL i=;i<n;i++)

         {

            LL k1=(i+n-)%n;

            mat[i][k1]=R;

            LL k2=(i+)%n;

            mat[i][k2]=L;

         }

     }

 }M_hxl,M_tom;

 Matrix Multiply(Matrix &x, Matrix &y)

 {

     LL i, j, k;

     Matrix z;

     z.ini();

     for(k = ; k < n; k ++)

         if(x.mat[][k])

         {

             for(j = ; j < n; j ++)

                 if(y.mat[k][j])

             z.mat[][j] = (z.mat[][j] + (LL)x.mat[][k] * y.mat[k][j]) % mod;

         }

     for(i = ; i < n; i ++)

     {

         z.mat[i][] = z.mat[i - ][n - ];

         for(j = ; j < n; j ++)

             z.mat[i][j] = z.mat[i - ][j - ];

     }

     return z;

 }

 void cs()

 {

     LL i,j;

     for(i=;i<n;i++)

     {

         printf("\n");

         for(j=;j<n;j++)

         printf("%I64d ",M_hxl.mat[i][j]);

     }

     printf("\n");

 }

 void power_sum2()

 {

     M_hxl.init();

     M_hxl.make_first();

     M_tom.init();

     cs();

     while(m)

     {

         if(m&)

         {

             M_tom=Multiply(M_tom,M_hxl);

         }

         m=m>>;

         M_hxl=Multiply(M_hxl,M_hxl);

     }

     for(LL i=;i<n;i++)

     {

         LL sum=;

         for(LL j=;j<n;j++)

         {

             sum=(sum+f[j]*M_tom.mat[i][j])%mod;

         }

         if(i==)printf("%I64d",sum);

         else printf(" %I64d",sum);

     }

     printf("\n");

 }

 int main()

 {

     LL T;

    while(scanf("%I64d",&T)>)

     {

         while(T--)

         {

             scanf("%I64d%I64d%I64d%I64d%I64d",&n,&m,&L,&R,&mod);

             for(LL i=;i<n;i++)

             scanf("%I64d",&f[i]);

             if(m==)

             {

                 for(LL i=;i<n;i++)

                 {

                     if(i==)printf("%I64d",f[i]%mod);

                     else printf(" %I64d",f[i]%mod);

                 }

                 printf("\n");

                 continue;

             }

             power_sum2();

         }

     }

     return ;

 }