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 c[201] = {0};
bool p[205] = {0};
void sieve() {
  long long i, j;
  j = 4;
  while (j <= 201) {
    p[j] = 1;
    j += 2;
  }
  long k = 1;
  for (i = 3; i <= (201); i += 2) {
    if (p[i] == 0) {
      j = i;
      while (j + i <= 201) {
        p[j + i] = 1;
        j += i;
      }
      if (p[i + 2] == 0) c[k++] = i;
    }
  }
}
long long npr(long n, long m) {
  long long ret = 1, i;
  for (i = 1; i <= m; i++) ret *= (n - i + 1);
  return ret;
}
long long ncr(long n, long m) {
  long long ret = 1, i;
  for (i = 1; i <= m; i++) {
    ret *= (n - i + 1);
    ret /= i;
  }
  return ret;
}
long long fact(long n) {
  long long ret = 1, i;
  for (i = 1; i <= n; i++) ret *= i;
  return ret;
}
long long gcd(long n, long m) {
  while (m > 0) {
    n = n % m;
    n = n ^ m;
    m = m ^ n;
    n = n ^ m;
  }
  return n;
}
long long lcm(long n, long m) {
  long long x = (n * m) / gcd(n, m);
  return x;
}
long is_prime(long n) {
  long long ii;
  if (n == 1) return 0;
  if (n == 2) return 1;
  if (n % 2 == 0) return 0;
  for (ii = 3; ii * ii <= n; ii = ii + 2)
    if (n % ii == 0) return 0;
  return 1;
}
double area_triangle(double x1, double y1, double x2, double y2, double x3,
                     double y3) {
  double a;
  a = x1 * (y2 - y3) + y1 * (x3 - x2) + ((x2 * y3) - (y2 * x3)) + 1e-9;
  a = a / 2 + 1e-9;
  return a;
}
double dist_2point(double x1, double y1, double x2, double y2) {
  double d;
  d = sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)));
  return d;
}
int binary_search(int a[], int n, int l, int u) {
  int mid;
  if (l > u) return 0;
  mid = floor(l + u) / 2;
  if (a[mid] == n)
    return mid;
  else if (a[mid] > n)
    binary_search(a, n, l, mid - 1);
  else
    binary_search(a, n, mid + 1, u);
}
bool is_palindrome(const string &s, int start, int end) {
  int siz = end - start + 1;
  if (siz % 2 == 0) {
    int middle = start + siz / 2;
    int i;
    for (i = 0; middle - 1 - i >= 0 && middle + i <= end &&
                s[middle - 1 - i] == s[middle + i];
         i++)
      ;
    if (middle - i - 1 < 0 || middle + i > end) return true;
  } else {
    int middle = start + siz / 2;
    int i;
    for (i = 1;
         middle - i >= 0 && middle + i <= end && s[middle - i] == s[middle + i];
         i++)
      ;
    if (middle - i < 0 || middle + i > end) return true;
  }
  return false;
}
int main() {
  long n, k;
  while (~scanf("%ld %ld", &n, &k)) {
    long i, a[101] = {0}, j, b = 1, l;
    for (i = 0; i < k; i++) scanf("%ld", &a[i]);
    long f = 0, c = 0, nn;
    nn = n;
    for (i = 0; i < k; i++) {
      printf("%ld ", a[i]);
      for (j = 0; j < n - 1; j++) {
        b = 0;
        c++;
        for (l = 0; l < k; l++)
          if (c == a[l]) {
            b = 1;
            break;
          }
        if (b == 1)
          n++;
        else
          printf("%ld ", c);
      }
      n = nn;
      printf("\n");
      f++;
    }
  }
  return 0;
}