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() {
  int t;
  t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    long long int arr[n + 1];
    long long int ans[n + 1];
    unordered_map<long long int, long long int> mp;
    arr[0] = 0;
    for (__typeof(n + 1) i = (1) - ((1) > (n + 1));
         i != (n + 1) - ((1) > (n + 1)); i += 1 - 2 * ((1) > (n + 1))) {
      cin >> arr[i];
      mp[arr[i]] = 1;
    }
    long long int p = 0;
    for (__typeof(n + 1) i = (1) - ((1) > (n + 1));
         i != (n + 1) - ((1) > (n + 1)); i += 1 - 2 * ((1) > (n + 1))) {
      if (arr[i] != arr[i - 1]) {
        ans[i] = arr[i - 1];
        mp[arr[i - 1]] = 1;
        continue;
      }
      while (mp[p]) {
        p++;
      }
      ans[i] = p;
      mp[p] = 1;
    }
    for (__typeof(n + 1) i = (1) - ((1) > (n + 1));
         i != (n + 1) - ((1) > (n + 1)); i += 1 - 2 * ((1) > (n + 1)))
      cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}
