Given an array a of length n, find another array, b, of length n such that:

  * for each i (1 ≤ i ≤ n) MEX(\\{b_1, b_2, …, b_i\})=a_i. 



The MEX of a set of integers is the smallest non-negative integer that doesn't belong to this set.

If such array doesn't exist, determine this.

Input

The first line contains an integer n (1 ≤ n ≤ 10^5) — the length of the array a.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ i) — the elements of the array a. It's guaranteed that a_i ≤ a_{i+1} for 1≤ i < n.

Output

If there's no such array, print a single line containing -1.

Otherwise, print a single line containing n integers b_1, b_2, …, b_n (0 ≤ b_i ≤ 10^6)

If there are multiple answers, print any.

Examples

Input


3
1 2 3


Output


0 1 2 

Input


4
0 0 0 2


Output


1 3 4 0 

Input


3
1 1 3


Output


0 2 1 

Note

In the second test case, other answers like [1,1,1,0], for example, are valid.
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  int ans[n];
  memset(ans, -1, sizeof(ans));
  int p = 0, cur = a[0], no = 0;
  if (a[0] == 1)
    ans[0] = 0;
  else if (a[0] != 0)
    no = 1;
  for (int i = 0; i < n; i++) {
    if (a[i] != cur) {
      ans[i] = cur;
      for (int j = cur + 1; j < a[i]; j++) {
        while (ans[p] != -1) p++;
        if (p > i) {
          no = 1;
          break;
        }
        ans[p] = j;
      }
      cur = a[i];
    }
    if (no) {
      break;
    }
  }
  if (no) {
    cout << "-1\n";
    return 0;
  }
  int mx = n + 2;
  for (int i = 0; i < n; i++) {
    if (ans[i] == -1) ans[i] = mx++;
    cout << ans[i] << " ";
  }
  return 0;
}
