Given a n × n matrix A and a positive integer k, find the sum S = A + A² + A³ + … + A^k.

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 10⁹) and m (m < 10⁴). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output the elements of S modulo m in the same way as A is given.

/*

矩阵乘法经典+二分

Sk=A+A2+A3+...+Ak

  =(1+Ak/2)*(A+A2+A3+...+Ak/2)+{Ak}

  =(1+Ak/2)*(Sk/2)+{Ak}

当k为偶数时不要大括号里面的数

*/

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

int n,m,k;

struct matrix

{

    int a[][];

    void init()

    {

        memset(a,,sizeof a);

        for(int i=;i<;i++) a[i][i]=;

    }

};

void print(matrix s)

{

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

    {

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

        {

            if (j)

                printf(" ");

            printf("%d",s.a[i][j]%m);

        }

        printf("\n");

    }

}

matrix m_add(matrix a,matrix b)//加法

{

    matrix c;

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

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

        c.a[i][j]=((a.a[i][j]+b.a[i][j])%m);

    return c;

}

matrix m_mul(matrix a,matrix b)//乘法

{

    matrix c;

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

    {

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

        {

            c.a[i][j]=;

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

              c.a[i][j]+=((a.a[i][k]*b.a[k][j])%m);

            c.a[i][j]%=m;

        }

    }

    return c;

}

matrix mul(matrix s,int k)//矩阵快速幂

{

    matrix ans;ans.init();

    while(k>=)

    {

        if(k&) ans=m_mul(ans,s);

        k>>=;

        s=m_mul(s,s);

    }

    return ans;

}

matrix sum(matrix s,int k)//矩阵前k项求和

{

    if(k==) return s;

    matrix tmp;tmp.init();

    tmp=m_add(tmp,mul(s,k>>));//计算1+A^(k/2)

    tmp=m_mul(tmp,sum(s,k>>));//计算(1+A^(k/2))*(A+A^2+A^3+...+A^(k/2))

    if(k&) tmp=m_add(tmp,mul(s,k));

    return tmp;

}

int main()

{

    while(cin>>n>>k>>m)

    {

        matrix s;

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

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

            scanf("%d",&s.a[i][j]);

        s=sum(s,k);

        print(s);

    }

}