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 OO = (int)2e9;
const double eps = 1e-9;
bool vis[1001];
int main() {
  int i, j, n, k, x;
  cin >> n >> k;
  vector<vector<int> > V(k + 1);
  for ((i) = (1); (i) < (k + 1); (i)++) {
    cin >> x;
    vis[x] = 1;
    V[i].push_back(x);
  }
  for ((i) = (1); (i) < (k + 1); (i)++) {
    if (((int)(V[i]).size()) == n) continue;
    for ((j) = (1); (j) < (n * k + 1); (j)++) {
      if (!vis[j]) vis[j] = 1, V[i].push_back(j);
      if (((int)(V[i]).size()) == n) break;
    }
  }
  for ((i) = (1); (i) < (k + 1); (i)++)
    for ((j) = 0; (j) < (n); (j)++) cout << V[i][j] << " \n"[j == n - 1];
  return 0;
}