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;
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
T cube(T x) {
  return x * x * x;
}
int main() {
  int n = ({
    int t;
    scanf("%d", &t);
    t;
  }),
      i = 0, p = ({
               int t;
               scanf("%d", &t);
               t;
             }),
      k = 0;
  set<int> b1;
  set<int>::iterator g1;
  int a[100], pos = 0, count = 1;
  ;
  for (int i = 0; i < (int)(p); i++) {
    a[i] = ({
      int t;
      scanf("%d", &t);
      t;
    });
    b1.insert(a[i]);
  }
  for (i = 0; i < p; i++) {
    printf("%d ", a[i]);
    count = 1;
    if (n != 1) {
      for (k = pos + 1;; k++) {
        g1 = b1.find(k);
        if (g1 != b1.end())
          continue;
        else {
          printf("%d ", k);
          count++;
        }
        if (count == n) break;
      }
    }
    printf("\n");
    ;
    pos = k;
  }
  return 0;
}