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;
void solve() {
  long long int n, temp = 0, prev = 0;
  cin >> n;
  long long int a[n];
  vector<long long int> ans;
  set<long long int> s;
  for (long long int i = 0; i < n; i += 1) {
    cin >> a[i];
    s.insert(a[i]);
  }
  for (long long int i = 0; i < n; i += 1) {
    if (a[i] != prev) {
      ans.push_back(prev);
      temp++;
      prev = temp;
      continue;
    }
    if (a[i] != temp) {
      if (!s.count(temp)) {
        ans.push_back(temp);
        temp++;
      } else {
        temp++;
        ans.push_back(temp);
        temp++;
      }
    } else {
      temp++;
      while (1) {
        if (!s.count(temp)) {
          ans.push_back(temp);
          temp++;
          break;
        }
      }
    }
  }
  for (long long int i = 0; i < ans.size(); i += 1) cout << ans[i] << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    solve();
    cout << '\n';
  }
  return 0;
}
