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;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return b * a / gcd(a, b);
}
long long int abs1(long long int a) {
  if (a < 0)
    return -1 * a;
  else
    return a;
}
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int fast_pow(long long base, long long n) {
  if (n == 0) return 1;
  if (n == 1) return base;
  long long halfn = fast_pow(base, n / 2);
  if (n % 2 == 0)
    return (halfn * halfn);
  else
    return (((halfn * halfn)) * base);
}
long long int findMMI_fermat(long long int n, long long int M) {
  return fast_pow(n, M - 2);
}
const double PI = 3.14159265358979323846;
int main() {
  map<int, int> ma;
  int n, k;
  cin >> n >> k;
  int a[k];
  for (int i = 0; i < k; i++) {
    cin >> a[i];
    ma[a[i]]++;
  }
  int kk = 1;
  for (int i = 0; i < k; i++) {
    cout << a[i] << " ";
    int j = 1;
    while (j < n) {
      if (ma[kk] == 0) {
        cout << kk << " ";
        kk++;
        j++;
      } else
        kk++;
    }
    cout << '\n';
  }
  return 0;
}