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.*;
public class dividing_org {

		 public static void main(String args[])
	    {
	        Scanner in = new Scanner(System.in);
	        int n = in.nextInt();
	        int k = in.nextInt();
	        int a[]= new int[k];
	        int arr[]=  new int[k];
	        for(int i=0 ; i<k ; i++) 
	        {
	            int x = in.nextInt();
	            arr[i]=a[i]=x;
	        }
	        Arrays.sort(arr);
	        int x = 1;
	       for(int i=0 ; i<k ; i++)
	       {
	           for(int j=0 ; j<n ; j++)
	           {
	                if(j==0)
	                 {
	                   System.out.print(a[i]+" ");
	                   if(Arrays.binarySearch(arr,x)>=0)
	                    x++;
	                 }
	                else
	                {
	                    if(Arrays.binarySearch(arr,x)<0)
	                    {
	                        System.out.print(x+" ");
	                        x++;
	                    }
	                    else
	                    {
	                        while(Arrays.binarySearch(arr,x)>=0)
	                         x++;
	                         System.out.print(x+" ");
	                         x++;
	                    }
	                }
	           //    System.out.println(x+" "+i+" "+j);
	               
	           }
	           System.out.println();
	       }
	    }
	}