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 

import java.util.Arrays;
import java.util.*;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author Dell
 */
public class NewMain46 {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int k = s.nextInt();

        List<Integer> q = new LinkedList<>();
        List<Integer> q1 = new LinkedList<>();
        int x = (n * k);
        for (int i = 1; i <= x; i++) {
            q.add(i);
        }
        for (int i = 0; i < k; i++) {
            int v = s.nextInt();
            //q.set(v - 1, 0);
            q.remove(Integer.valueOf(v));
            q1.add(v);
        }
        //  System.out.println(q1);
        int c = 0, f = 0;
        if (n == 1) {
            n++;
        }
        for (int i = 0; i < Math.max(q.size(), q1.size()); i++) {
            if (q.size() > 0) {
                System.out.print(q.get(i) + " ");
            }
            if ((i + 1) % n == 0 && f == k) {
                System.out.println("");
            } else if ((i + 1) % (n - 1) == 0 && f < k) {
                System.out.println(q1.get(f++) + " ");
            }

        }
   
        // System.out.println("");
    }

}