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.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.Queue;
import java.util.Scanner;

public class Main {	
	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n, k;
		n = in.nextInt();
		k = in.nextInt();
		int tmp = n *k;
		int a[] = new int[k];
		boolean fr[] = new boolean[tmp+1];
		
		for (int i = 0; i < k; ++i){
			a[i] = in.nextInt();	
			fr[a[i]] = true;
		}		
		tmp = 1;
		for (int i = 0; i < k; ++i){			
			for (int j = 0; j < n-1; ++j){
				while (fr[tmp]) 
					++tmp;
				out.printf("%d ", tmp);
				fr[tmp] = true;
			}
			out.printf("%d\n", a[i]);
		}
		in.close();
		out.close();
	}
}