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 M = 1000000007, BIG = 0x3f3f3f3f;
int n, k;
vector<int> a;
vector<vector<int> > Process() {
  vector<vector<int> > ans(k, vector<int>());
  vector<bool> vis(n * k + 5, 0);
  for (int i = 0; i < k; i++) {
    ans[i].push_back(a[i]);
    vis[a[i]] = 1;
  }
  int cur = 1;
  for (int i = 0; i < k; i++) {
    for (int j = 1; j < n; j++, cur++) {
      while (vis[cur]) cur++;
      ans[i].push_back(cur);
    }
  }
  return ans;
}
void Output() {
  const vector<vector<int> > ans = Process();
  for (int i = 0; i < ans.size(); i++) {
    for (int j = 0; j < ans[i].size(); j++) cout << ans[i][j] << " ";
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  while (cin >> n) {
    cin >> k;
    a.resize(k);
    for (int i = 0; i < k; i++) cin >> a[i];
    Output();
  }
  return 0;
}