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;
double PI = 3.1415926535897932384626;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  set<long long> ss;
  for (long long i = 0; i < arr[n - 1]; i++) {
    ss.insert(i);
  }
  vector<long long> ans;
  for (long long i = n - 1; i >= 1; i--) {
    if (arr[i] != arr[i - 1]) {
      ans.emplace_back(arr[i - 1]);
      if (ss.size() > 1 && ss.find(arr[i - 1]) != ss.end())
        ss.erase(ss.find(arr[i - 1]));
    } else {
      if (i == arr[i]) {
        ans.emplace_back(arr[n - 1] + 1);
        continue;
      }
      ans.emplace_back(*(ss.rbegin()));
      auto it = ss.end();
      if (ss.size() > 1) ss.erase(prev(it));
    }
  }
  ans.emplace_back(*(ss.rbegin()));
  reverse(ans.begin(), ans.end());
  for (auto &noo : ans) {
    cout << noo << " ";
  }
  return 0;
}
