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 JAVA (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int l = sc.nextInt();

        int A[][] = new int[n][m];
        int B[][] = new int[m][l];
        int cnt = 0, ai = 0, bi = 0, abj = 0, x = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int num = sc.nextInt();
                A[i][j] = num;
            }
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < l; j++) {
                int num = sc.nextInt();
                B[i][j] = num;
            }
        }
        while (x < n * l) {
            cnt += A[ai][abj] * B[abj][bi];
            if (abj + 1 == m) {
                System.out.print(cnt);
                x++;
                if (x % n != 0) {
                    System.out.print(" ");
                } else {
                    System.out.println("");
                }
                cnt = 0;
                abj = 0;
                if (bi < l - 1) {
                    bi++;
                } else {
                    bi = 0;
                    if (ai < n - 1) {
                        ai++;
                    }
                }
            }else{
                abj++;
            }
        }
    }
}