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() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<int> need(n + 1, 0);
  for (int i = 0; i < n; i++) {
    if (a[i] > i + 1) {
      cout << -1 << endl;
      return;
    }
    need[a[i]]++;
  }
  vector<int> ans(n, -1);
  int where = 0;
  for (int i = 0; i < n; i++) {
    if (ans[i] == -1) {
      while (where < n && need[where]) {
        where++;
      }
      ans[i] = where;
      if (where < n) {
        where++;
      }
    }
    need[a[i]]--;
    if (need[a[i]] == 0 && i + 1 < n) {
      ans[i + 1] = a[i];
    }
  }
  for (int& x : ans) {
    cout << x << " ";
  }
  cout << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  srand(chrono::steady_clock::now().time_since_epoch().count());
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
