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.Scanner;

public class Main {
    private static Scanner scanner = new Scanner(System.in);
    private static int OO = (int) 1e5 + 9;
    private static int[] dx = {0, 0, 1, -1};
    private static int[] dy = {1, -1, 0, 0};
    private static int[] num = new int[OO];

    public static void main(String[] args) {
        int n = scanner.nextInt(), k = scanner.nextInt(), o = n * k;
        int[] arr = new int[k];
        boolean[] visited = new boolean[o + 1];
        for (int i = 0; i < k; i++) {
            arr[i] = scanner.nextInt();
            visited[arr[i]] = true;
        }
        for (int i = 0; i < k; i++) {
            System.out.print(arr[i]);
            int a = n - 1, b = 1;
            while (a > 0) {
                if (visited[b]) {
                    b += 1;
                } else {
                    System.out.print(" "+b);
                    visited[b] = true;
                    a--;
                }
            }
            System.out.println(" ");
        }
    }
}