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;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto h1 = hash<T1>{}(p.first);
    auto h2 = hash<T2>{}(p.second);
    return h1 ^ h2;
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int n;
  cin >> n;
  vector<long long int> v(n), ans(n, -1);
  unordered_map<long long int, long long int> mp;
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
    mp[v[i]]++;
  }
  unordered_map<long long int, short int> mp2;
  mp2[v[n - 1]] = true;
  for (long long int i = n - 2; i >= 0; i--) {
    if (mp2[v[i]] == 0) {
      ans[i + 1] = v[i];
      mp2[v[i]] = 1;
      mp[v[i]]++;
    }
  }
  set<long long int> used;
  long long int x = 0;
  for (long long int i = 0; i < n; i++) {
    while (mp[x]) x++;
    if (ans[i] == -1) {
      ans[i] = x;
      mp[x]++;
      mp[v[i]]--;
    }
  }
  for (auto h : ans) cout << h << " ";
  return 0;
}
