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;
string to_string(long long n) {
  stringstream s;
  s << n;
  return s.str();
}
long long to_long(string s) {
  stringstream ss;
  long long x;
  ss << s;
  ss >> x;
  return x;
}
long long gcd(long long n, long long m) { return m == 0 ? n : gcd(m, n % m); }
bool b[100000];
bool c[100000];
int main() {
  int n, k;
  cin >> n;
  cin >> k;
  int a[k + 1];
  for (int i = 1; i <= 100000; i++) {
    b[i] = false;
    c[i] = false;
  }
  for (int i = 1; i <= k; i++) {
    cin >> a[i];
    b[a[i]] = true;
  }
  for (int i = 1; i <= k; i++) {
    cout << a[i];
    cout << " ";
    int count1 = 1;
    for (int j = 1;; j++) {
      if (b[j] == false && c[j] == false) {
        cout << j;
        cout << " ";
        count1++;
        c[j] = true;
      }
      if (count1 == n) {
        cout << endl;
        break;
      }
    }
  }
  return 0;
}