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 n;
  cin >> n;
  bool can = true;
  vector<long long> v(n);
  set<long long> s;
  vector<long long> change;
  map<long long, long long> mp;
  for (long long i = 1; i <= n; i++) {
    cin >> v[i - 1];
    mp[v[i - 1]]++;
    if (i > 1) {
      if (v[i - 1] != v[i - 2]) {
        change.push_back(i - 1);
      }
    }
    s.insert(v[i - 1]);
    if ((v[i - 1]) > i) {
      can = false;
      break;
    }
  }
  if (can) {
    long long i = 0;
    long long count = 0;
    for (long long z = 0; z <= n + 1; z++) {
      if (mp[z] == 0) {
        cout << z << " ";
        count++;
        if (count == n) {
          break;
        }
        mp[v[i]]--;
        while (mp[v[i]] == 0) {
          cout << v[i] << " ";
          if (z <= v[i]) {
            z++;
          }
          count++;
          if (count == n) {
            break;
          }
          i++;
          mp[v[i]]--;
        }
        if (count == n) {
          break;
        }
        i++;
      }
    }
  } else {
    cout << -1 << endl;
  }
}
