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 Scala (judge verdict: Wrong Answer)

object Main extends App {

  val Array(n, m, l) = io.StdIn.readLine().split(" ").map(_.toInt)

  val a = Array.ofDim[Int](n,m)
  val b = Array.ofDim[Int](m,l)
  for(i <- 0 until n)
    a(i) = io.StdIn.readLine().split(" ").map(_.toInt)
  for(i <- 0 until m)
    b(i) = io.StdIn.readLine().split(" ").map(_.toInt)

  val ans: Array[Array[Long]] = Array.ofDim[Long](n,l)
  for(i <- 0 until n; j <- 0 until l)
    ans(i)(j) = (0 until m).foldLeft(0) {
      (acc,k) => acc + a(i)(k) * b(k)(j)
    }

  val sb = StringBuilder.newBuilder
  for(e <- ans)
    sb.append(e.mkString(" ")).append("\n")
  print(sb)
}