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>
const long long max9 = 10 + 1e9, max6 = 10 + 1e6, max12 = 10 + 1e12,
                max15 = 10 + 1e15;
const long long min6 = -1 * max6, min9 = -1 * max9, min12 = -1 * max12,
                min15 = -1 * max15;
const long long R = 7 + 1e9, NN = 10 + 1e5;
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  bool a[n * k + 1];
  memset(a, false, sizeof(a));
  int t[k];
  for (int i = (0); i < (k); ++i) cin >> t[i];
  for (int i = (0); i < (k); ++i) {
    a[t[i] - 1] = true;
  }
  for (int i = (0); i < (k); ++i) {
    cout << t[i] << " ";
    int o = 1;
    for (int j = (0); j < (n * k + 1); ++j) {
      if (!a[j]) {
        cout << j + 1 << " ";
        o++;
        a[j] = true;
      }
      if (o == n) {
        cout << endl;
        break;
      }
    }
  }
  return 0;
}