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;
template <typename T>
T bitCount(T x) {
  T c = 0;
  while (x) {
    x &= x - 1;
    c++;
  }
  return c;
}
void A_Dividing_Orange() {
  int n, k, j;
  bool taken[1000];
  int a[100];
  cin >> n >> k;
  memset(taken, 0, sizeof(taken));
  for (auto i = ((k) - (k)); (i) < (k); (i)++) {
    cin >> a[i];
    taken[a[i]] = true;
  }
  j = 1;
  for (auto ii = ((k) - (k)); (ii) < (k); (ii)++) {
    cout << a[ii] << " ";
    for (auto i = (1); (i) != (n); (i)++) {
      if (i - 1) cout << " ";
      while (taken[j]) j++;
      cout << j;
      j++;
    }
    cout << endl;
  }
}
int main() {
  A_Dividing_Orange();
  return 0;
}