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 toi(string s) {
  istringstream is(s);
  long long x;
  is >> x;
  return x;
}
int conversion(string p) {
  int num;
  num = atoi(p.c_str());
  return num;
}
string toString(int h) {
  stringstream ss;
  ss << h;
  return ss.str();
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return (a * (b / gcd(a, b))); }
bool cmp(pair<int, int> a, pair<int, int> b) { return a.first > b.first; }
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> v(k);
  bool vis[(n * k) + 1];
  memset(vis, false, sizeof vis);
  for (int i = 0; i < k; ++i) {
    cin >> v[i];
    vis[v[i]] = true;
  }
  for (int i = 0; i < k; ++i) {
    cout << v[i] << " ";
    int c = 1;
    for (int j = 1; j <= n * k; ++j) {
      if (c == n) continue;
      if (!vis[j]) {
        cout << j << " ";
        vis[j] = true;
        c++;
      }
    }
    cout << endl;
  }
}
int main() {
  solve();
  return 0;
}