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.lang.*;
import java.io.*;

public class Ideone
{
	public static void main (String[] args) throws java.lang.Exception
{
	Scanner in = new Scanner(System.in);
	int n, k,flag=0,l=0;
	n=in.nextInt();
	k=in.nextInt();
	int a[]=new int[k];
	int output[]=new int[n];
	for(int i=0;i<k;i++)
	{
		a[i]=in.nextInt();
	}
	for(int count=1;count <=n*k;)
	{
		output[0]=a[l];
		System.out.print(output[0] + " ");
		for(int i=1;i<n;)
		{
			for(int j=0;j<k;j++)
			{
				if(count==a[j])
				{
					count++;
					flag++;
					break;
				}
			}
			if(flag==0)
			{
				output[i] = count;
				System.out.print(output[i] + " ");
				i++;
				count++;
			}
			flag=0;
		}
		System.out.println(" ");
		l++;
		if(l==k)
		break;
	}
}
}