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;
bool isSorted(string s) {
  for (long long int i = 0; i + 1 < s.size(); i++) {
    if (s[i] > s[i + 1]) {
      return false;
    }
  }
  return true;
}
bool isPrime(long long int n) {
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long int stringToINT(string s) {
  long long int sum = 0;
  for (long long int j = 0; j < s.size(); j++) {
    sum = sum * 10 + (s[j] - '0');
  }
  return sum;
}
bool isINT(string s) {
  for (long long int i = 0; i < s.size(); i++) {
    if (s[i] < '0' || s[i] > '9') {
      return false;
    }
  }
  return true;
}
vector<long long int> preSum(vector<long long int> v) {
  vector<long long int> sum;
  sum.push_back(v[0]);
  for (long long int i = 1; i < v.size(); i++) {
    sum.push_back(v[i] + sum[i - 1]);
  }
  return sum;
}
vector<string> subString(vector<string> v, string s) {
  for (long long int i = 0; i < s.size(); i++)
    for (long long int len = 1; len <= s.size() - i; len++)
      v.push_back(s.substr(i, len));
  return v;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k, x;
  cin >> n >> k;
  bool dn[(n * k) + 1];
  memset(dn, false, sizeof(dn));
  queue<long long int> q;
  vector<long long int> v[k];
  for (long long int i = 0; i < k; i++) {
    cin >> x;
    v[i].push_back(x);
    dn[x] = true;
  }
  for (long long int i = 1; i <= n * k; i++) {
    if (dn[i] == false) q.push(i);
  }
  for (long long int i = 0; i < k; i++) {
    for (long long int j = 1; j < n; j++) {
      v[i].push_back(q.front());
      q.pop();
    }
  }
  for (long long int i = 0; i < k; i++) {
    for (long long int j = 0; j < v[i].size(); j++) {
      cout << v[i][j] << " ";
    }
    cout << endl;
  }
  return 0;
}