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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class A {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer sc = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(sc.nextToken());
        int k = Integer.parseInt(sc.nextToken());
        int[] a = new int[k];
        boolean[] b = new boolean[n * k];
        sc = new StringTokenizer(br.readLine());
        for (int i = 0; i < k; i++) {
            a[i] = Integer.parseInt(sc.nextToken()) - 1;
            b[a[i]] = true;
        }
        int counter;
        for (int i = 0; i < k; i++) {
            counter = 0;
            for (int j = 0; j < n * k && counter < n - 1; j++) {
                if (b[j] == false) {
                    counter++;
                    b[j] = true;
                    System.out.print((j + 1) + " ");
                }
            }
            System.out.println(a[i] + 1);
        }
    }
}