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 freq[1000001] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    int arr[n];
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      freq[arr[i]]++;
    }
    for (int i = 0; i < n; i++) {
      if (arr[i] > i + 1) {
        cout << "-1";
        exit(0);
      }
    }
    int ind;
    if (arr[0] == 0) {
      ind = 1;
      cout << "1 ";
      ind++;
    } else {
      ind = 0;
      cout << "0 ";
      ind++;
    }
    freq[arr[0]]--;
    for (int i = 1; i < n; i++) {
      if (freq[arr[i - 1]] == 0) {
        cout << arr[i - 1] << " ";
        if (ind == arr[i - 1]) ind++;
      } else {
        while (freq[ind] != 0) {
          ind++;
        }
        cout << ind << " ";
        ind++;
      }
      freq[arr[i]]--;
    }
  }
}
