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 bgm(long long a, long long b, long long m) {
  a %= m;
  long long rem = 1;
  while (b) {
    if (b & 1) rem = (rem * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return rem;
}
long long inv(long long a, long long mod) { return bgm(a, mod - 2, mod); }
long long in() {
  long long a;
  assert(scanf("%lld", &a) != EOF);
  return a;
}
double dl() {
  double a;
  assert(scanf("%lf", &a) != EOF);
  return a;
}
const int MAX = 35;
const int LEN = 13;
const long long MOD = 1000000007;
double ttm;
int n, k;
int x;
vector<int> adj[MAX];
int cnt[MAX * MAX];
int main() {
  n = in(), k = in();
  for (int i = 1; i <= k; i++) {
    x = in();
    adj[i].push_back(x);
    cnt[x] = 1;
  }
  for (int i = 1; i <= n * k; i++) {
    if (cnt[i] == 1) continue;
    for (int j = 1; j <= k; j++) {
      if (((long long)((adj[j]).size())) < n) {
        adj[j].push_back(i);
        break;
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= n; j++) {
      printf("%lld", (long long)(adj[i][j - 1])), printf(" ");
    }
    puts("");
  }
  return 0;
}