Problem:
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Matrix Multiplication

Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:

\[
c_{ij} = \sum_{k=1}^m a_{ik}b_{kj}
\]

where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.

Input

In the first line, three integers $n$, $m$ and $l$ are given separated by space characters

In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.

Output

Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.

Constraints

$1 \leq n, m, l \leq 100$

$0 \leq a_{ij}, b_{ij} \leq 10000$

Sample Input

3 2 3
1 2
0 3
4 5
1 2 1
0 3 2

Sample Output

1 8 5
0 9 6
4 23 14

Note

      解説

Solution in C++11 (judge verdict: Accepted)

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<cstdlib>
#include<iomanip>
#include<queue>
#include<set>
#include <valarray>
#include<stack>
#include<sstream>
#include<math.h>

using namespace std;

void Solution()
{
    int n, m, l;
    long int a[100][100] = {{0}};
    long int b[100][100] = {{0}};
    //long int c[100][100] = {{0}};
    long int sum = 0;
    cin>>n>>m>>l;
    
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            cin>>a[i][j];
        }
    }
    
     for(int i = 0; i < m; i++)
    {
        for(int j = 0; j < l; j++)
        {
            cin>>b[i][j];
        }
    }
    
     for(int i = 0; i < n; i++)
    {
         for(int x = 0; x < l; x++)
         {
            for(int j = 0; j < m; j++)
            {
               sum += a[i][j] * b[j][x];
            }
             if(x == l - 1)
             {
                 cout<<sum;
             }
             else
             {
                cout<<sum<<" ";
             }
             sum = 0;
         }
         
         cout<<endl;
    }
}

int main()
{
    Solution();
    return 0;
}