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.Arrays;
import java.util.Scanner;

public class Main
{
	static Scanner scan=new Scanner(System.in);
	
	public static void main(String[] args)
	{
		int n=scan.nextInt(),k=scan.nextInt(),temp=n*k;
		int a[]=new int[temp+1];
		int ans[][]=new int[k][n];
		Arrays.fill(a,1);
		for(int i=1;i<=k;i++)
		{
			ans[i-1][0]=scan.nextInt();
			a[ans[i-1][0]]--;
		}
		if(temp!=k)
		{
			int pop=1;
			for(int i=0;i<k;i++)
				for(int j=1;j<n;j++)
					if(a[pop]!=0)
						ans[i][j]=pop++;
					else {pop++;j--;}
		}
		for(int i=0;i<k;i++)
		{
			for(int j=0;j<n;j++)
				System.out.print(ans[i][j]+" ");
			System.out.println();
		}
	}
}