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>
using namespace std;
const int N = (int)2e5 + 3;
const int INF = 1e9 + 7;
int n, k, used[N], a[N], used2[N];
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= k; ++i) {
    scanf("%d", a + i);
    used2[a[i]] = 1;
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (j == 1) {
        printf("%d ", a[i]);
        used[a[i]] = 1;
      } else {
        for (int x = 1; x <= n * k; ++x) {
          if (!used[x] && !used2[x]) {
            used[x] = 1;
            printf("%d ", x);
            break;
          }
        }
      }
    }
    printf("\n");
  }
  return 0;
}