Problem: One day Ms Swan bought an orange in a shop. The orange consisted of n·k segments, numbered with integers from 1 to n·k. 

There were k children waiting for Ms Swan at home. The children have recently learned about the orange and they decided to divide it between them. For that each child took a piece of paper and wrote the number of the segment that he would like to get: the i-th (1 ≤ i ≤ k) child wrote the number ai (1 ≤ ai ≤ n·k). All numbers ai accidentally turned out to be different.

Now the children wonder, how to divide the orange so as to meet these conditions:

  * each child gets exactly n orange segments; 
  * the i-th child gets the segment with number ai for sure; 
  * no segment goes to two children simultaneously. 



Help the children, divide the orange and fulfill the requirements, described above.

Input

The first line contains two integers n, k (1 ≤ n, k ≤ 30). The second line contains k space-separated integers a1, a2, ..., ak (1 ≤ ai ≤ n·k), where ai is the number of the orange segment that the i-th child would like to get.

It is guaranteed that all numbers ai are distinct.

Output

Print exactly n·k distinct integers. The first n integers represent the indexes of the segments the first child will get, the second n integers represent the indexes of the segments the second child will get, and so on. Separate the printed numbers with whitespaces.

You can print a child's segment indexes in any order. It is guaranteed that the answer always exists. If there are multiple correct answers, print any of them.

Examples

Input

2 2
4 1


Output

2 4 
1 3 


Input

3 1
2


Output

3 2 1 

Solution (in JAVA):

Solution:

import java.util.*;
import java.math.*;

public class gfg
{
  public static void main(String args[])
  {
      Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      int k = sc.nextInt();
      int[] a = new int[k];
      
      for(int i=0 ; i<k ; i++)
         a[i] = sc.nextInt();
         
      int[] b = Arrays.copyOf(a, k);
      Arrays.sort(b);
         
      int total = n*k + 1; 
      int[] available = new int[total];
      int count = 0;
      
      for(int i=1 ; i<=total ; i++)
      {
          if(Arrays.binarySearch(b,i) < 0)
          {
              available[count] = i;
              count++;
          }
      }
      
      count = 0;
      for(int i=0 ; i<k ; i++)
      {
          System.out.print(a[i] + " ");
          for(int j=0 ; j<n-1 ; j++)
          {
              System.out.print(available[count++] + " ");
          }
          System.out.println("");
      }
  } 
}