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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws IOException {

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] strs = br.readLine().split(" ");
		int n = Integer.parseInt(strs[0]);
		int m = Integer.parseInt(strs[1]);
		int l = Integer.parseInt(strs[2]);

		int[][] list01 = new int[n][m];
		int[][] list02 = new int[m][l];
		double[][] list03 = new double[n][l];

		for (int i = 0; i < n; i++) {
			String[] str = br.readLine().split(" ");
			for (int j = 0; j < m; j++) {
				list01[i][j] += Integer.parseInt(str[j]);
			}
		}

		for (int i = 0; i < m; i++) {
			String[] str = br.readLine().split(" ");
			for (int j = 0; j < l; j++) {
				list02[i][j] += Integer.parseInt(str[j]);
			}
		}

		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				for (int k = 0; k < l; k++) {
					list03[i][k] += (double)list01[i][j] * list02[j][k];
				}
			}
		}

		for (int i = 0; i < n; i++) {
			for (int j = 0; j < (l - 1); j++) {
				System.out.print(list03[i][j] + " ");
			}
			System.out.println(list03[i][l - 1]);
		}

	}
}