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 (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){
  int i,j,k,n,m,l;

  scanf("%d %d %d",&n,&m,&l);

  if(n>=100) printf("Too Large\n");
  else if(m>=100) printf("Too Large\n");
  else if(l>=100) printf("Too Large\n");
  
  
  int A[n][m],B[m][l],C[n][l];

  for(i=0; i<n; i++)//A
    for(j=0; j<m; j++)
      scanf("%d",&A[i][j]);

  for(i=0; i<m; i++)//B
    for(j=0; j<l; j++)
      scanf("%d",&B[i][j]);

  for(i=0; i<n; i++)
    for(j=0; j<l; j++)
      C[i][j] = 0;

  /*Cを計算*/
  for(i=0; i<n; i++)
    for(j=0; j<l; j++)
      for(k=0; k<m; k++)
	C[i][j] += A[i][k] * B[k][j];

  for(i=0; i<n; i++){
    for(j=0; j<l; j++){
      if(j==0) printf("%d",C[i][j]);
      else printf(" %d",C[i][j]);
    }
    printf("\n");
  }

  return 0;
}