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 practice {
	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		int [] arr = new int[n*k+1];
		int [] per = new int[k];
		Arrays.fill(arr,0);
		for(int i = 0; i<k; i++) { 
			per[i] = sc.nextInt();
			arr[per[i]] = -1;
			
		}
		
		for(int i = 0; i<k; i++) {
			System.out.print(per[i] + " ");
			int count = 1;
			for(int j = 1; j<= n*k && count<n; j++) {
				if(arr[j] == 0) {
					System.out.print(j + " ");
					arr[j] = -1;
					count++;
				}
			}
			System.out.println();
		}
		
	}
}