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() {
  long long int n;
  cin >> n;
  vector<long long int> vec(n), copy(n);
  unordered_set<long long int> mp;
  for (long long int i = 0; i < n; i++) {
    cin >> vec[i];
    copy[i] = -1;
  }
  int flag = 0;
  for (long long int i = 1; i < n; i++) {
    if (vec[i] != vec[i - 1]) {
      copy[i] = vec[i - 1];
      mp.insert(vec[i - 1]);
    }
    if (vec[i] - i >= 2) {
      flag = 1;
      break;
    }
  }
  mp.insert(vec[n - 1]);
  if (flag == 1 || vec[0] >= 2) {
    cout << -1;
    cout << "\n";
    return 0;
  }
  int start = 0;
  for (int i = 0; i < n; i++) {
    if (copy[i] == -1) {
      while (mp.find(start) != mp.end()) {
        start++;
      }
      copy[i] = start;
    }
    mp.insert(start);
    cout << copy[i] << " ";
  }
  cout << "\n";
}
