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 MOD = 1e9 + 7;
const int N = 4e5 + 5;
int gcd(int n, int m) {
  if (n == 0)
    return m;
  else
    return gcd(m % n, n);
}
int binaryToDecimal(int n) {
  int num = n;
  int dec_value = 0;
  int base = 1;
  int temp = num;
  while (temp) {
    int last_digit = temp % 10;
    temp = temp / 10;
    dec_value += last_digit * base;
    base = base * 2;
  }
  return dec_value;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long n, k, a[32], i, j;
  set<long> s;
  cin >> n >> k;
  for (i = 1; i <= n * k; ++i) s.insert(i);
  for (i = 0; i < k; ++i) {
    cin >> a[i];
    s.erase(s.find(a[i]));
  }
  for (i = 0; i < k; ++i) {
    cout << a[i] << " ";
    for (j = 0; j < n - 1; ++j) {
      cout << *s.begin() << " ";
      s.erase(s.begin());
    }
    cout << endl;
  }
  return 0;
}