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 s_244A {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int N = s.nextInt(), K = s.nextInt();
        int[] aArr = new int[K];
        int[] allSegs = new int[N * K];
        for (int i = 0; i < N * K; i++) {
            allSegs[i] = i + 1;
        }
        for (int i = 0; i < K; i++) {
            aArr[i] = s.nextInt();
        }

        int x=0;
        for (int i = 0; i < K; i++) {
            System.out.print(aArr[i] + " ");
            for (int j = 1; j < N; j++) {
                while(isInArr(allSegs[x],aArr))
                {
                    x++;
                }
                System.out.print(allSegs[x++]+" ");
            }
            System.out.println();
        }

    }

    static boolean isInArr(int value, int[] arr)
    {
        for (int i = 0; i < arr.length; i++) {
            if(value==arr[i]) return true;
        }
        return false;
    }
}