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;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  long long arr[n + 5];
  long long ans[n + 5];
  long long present[100002];
  memset(present, 0, sizeof(present));
  for (long long x = 1; x <= n; x++) {
    cin >> arr[x];
    present[arr[x]] = 1;
  }
  long long limit = 0;
  for (long long x = 1; x <= n; x++) {
    if (x == 1) {
      if (arr[x] == 1) {
        ans[x] = 0;
        present[0] = 1;
      } else {
        for (long long y = 1; y <= 100000; y++) {
          if (present[y] == 0) {
            ans[x] = y;
            present[y] = 1;
            break;
          }
        }
      }
      continue;
    }
    if (arr[x] != arr[x - 1]) {
      ans[x] = arr[x - 1];
      present[ans[x]] = 1;
    } else {
      for (long long y = ans[x - 1]; y <= 100000; y++) {
        if (present[y] == 0) {
          ans[x] = y;
          present[y] = 1;
          break;
        }
      }
    }
  }
  for (long long x = 1; x <= n; x++) {
    cout << ans[x] << " ";
  }
  cout << "\n";
}
