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 CPP):

Solution:
#include <bits/stdc++.h>
int main(void) {
  int *orange, **gain, n, k, i, j, p;
  scanf("%d %d", &n, &k);
  orange = (int *)calloc(n * k, sizeof(int));
  gain = (int **)calloc(k, sizeof(int *));
  for (i = 0; i < k; i++) gain[i] = (int *)calloc(n, sizeof(int));
  for (i = 0; i < n * k; i++) orange[i] = 1;
  for (i = 0; i < k; i++) {
    scanf("%d", &gain[i][0]);
    orange[gain[i][0] - 1] = 0;
  }
  p = 0;
  for (i = 0; i < k; i++) {
    for (j = 1; j < n; j++) {
      do {
        if (orange[p]) {
          gain[i][j] = p + 1;
        }
      } while (++p && orange[p - 1] == 0);
    }
  }
  for (i = 0; i < k; i++) {
    for (j = 0; j < n; j++) {
      printf("%d", gain[i][j]);
      putchar((j == n - 1) ? '\n' : ' ');
    }
  }
  for (i = 0; i < k; i++) free(gain[i]);
  free(gain);
  free(orange);
  return 0;
}