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;
ifstream in("input.in");
ofstream out("output.out");
const long long nmx = 2e5 + 5;
const long long MOD = 1e9 + 7;
bool us[1000];
int main() {
  ios::sync_with_stdio(0);
  long long n, k, a[nmx];
  vector<int> ans[40];
  cin >> n >> k;
  for (long long i = 1; i <= k; ++i) {
    cin >> a[i];
    ans[i].push_back(a[i]);
    us[a[i]] = 1;
  }
  for (long long i = 1; i <= n * k; ++i)
    if (us[i] == 0) {
      long long mn = 1 << 30, wh = 0;
      for (long long j = 1; j <= k; ++j) {
        if (ans[j].size() < mn) {
          mn = ans[j].size();
          wh = j;
        }
      }
      ans[wh].push_back(i);
    }
  for (long long i = 1; i <= k; ++i) {
    for (auto k : ans[i]) cout << k << " ";
    cout << "\n";
  }
  return 0;
}