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 maxn = 40;
int n, k, lim;
int output[maxn][maxn];
int cnt[maxn];
int m[maxn * maxn];
void printans() {
  for (int i = 0; i < k; i++) {
    if (i > 0) cout << endl;
    for (int j = 0; j < n; j++) {
      if (j > 0) cout << ' ';
      cout << output[i][j] + 1;
    }
  }
}
int dfs(int cur) {
  if (cur == lim) {
    printans();
    return 1;
  }
  if (m[cur] != -1) {
    if (dfs(cur + 1)) return 1;
  } else {
    for (int i = 0; i < n; i++)
      if (cnt[i] < n) {
        output[i][cnt[i]++] = cur;
        if (dfs(cur + 1)) return 1;
        cnt[i]--;
      }
  }
  return 0;
}
int main() {
  cin >> n >> k;
  memset(m, -1, sizeof(m));
  memset(cnt, 0, sizeof(cnt));
  lim = n * k;
  for (int i = 0; i < k; i++) {
    int tmp;
    cin >> tmp;
    m[tmp - 1] = i;
    output[i][cnt[i]++] = tmp - 1;
  }
  int t = 0;
  for (int i = 0; i < lim; i++)
    if (m[i] == -1) {
      while (cnt[t] >= n) t++;
      output[t][cnt[t]++] = i;
    }
  printans();
  return 0;
}