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;
struct visit {
  int num;
  bool is_visit;
};
int A[100];
visit V[1000];
int Ck[100];
int n, k;
int Compare() {
  for (int j = (0), _b = ((n * k) - 1); j <= _b; j++) {
    if (V[j].is_visit == false) {
      V[j].is_visit = true;
      return V[j].num;
    }
  }
  return -1;
}
int main(void) {
  n = 0, k = 0;
  cin >> n >> k;
  for (int i = (0), _b = ((n * k) - 1); i <= _b; i++) {
    V[i].num = i + 1;
    V[i].is_visit = false;
  }
  for (int m = (0), _b = ((k)-1); m <= _b; m++) {
    cin >> A[m];
    for (int q = (0), _b = ((n * k) - 1); q <= _b; q++)
      if (A[m] == V[q].num) V[q].is_visit = true;
  }
  if (k == 1) {
    for (int i = ((n * k) - 1), _a = (0); i >= _a; i--) cout << i + 1 << " ";
    return 0;
  }
  for (int i = (0), _b = ((k)-1); i <= _b; i++) {
    Ck[0] = A[i];
    for (int g = (0), _b = ((n - 1) - 1); g <= _b; g++) Ck[g + 1] = Compare();
    sort(Ck, Ck + n);
    for (int s = (0), _b = ((n)-1); s <= _b; s++) {
      cout << Ck[s];
      if (s + 1 == n)
        cout << "\n";
      else
        cout << " ";
    }
  }
  return 0;
}