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;
long long int position[1000001];
long long int checker[1000001];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    position[a[i]]++;
  }
  vector<long long int> answer;
  queue<long long int> q;
  long long int temp = 0;
  for (long long int i = 0; i < n; i++) {
    if (position[i] == 0) {
    } else {
      while (position[i] != 0) {
        if (q.size() == 1) {
          long long int f = q.front();
          position[i]--;
          answer.push_back(f);
          q.pop();
        } else {
          while (position[temp] || checker[temp]) temp++;
          checker[temp]++;
          position[i]--;
          answer.push_back(temp++);
        }
      }
      q.push(i);
    }
  }
  for (auto it = answer.begin(); it != answer.end(); it++) {
    cout << *it << " ";
  }
  cout << "\n";
  return 0;
}
