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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int arr[100005], brr[100005], fre[100005];
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i];
  }
  memset(brr, -1, sizeof(brr));
  for (int i = 1; i <= n; ++i) {
    if (arr[i] != arr[i - 1]) {
      brr[i] = arr[i - 1];
      fre[brr[i]] = 1;
    }
  }
  fre[brr[n]] = 1;
  int m = 0;
  for (int i = 1; i <= n; ++i) {
    while (fre[m]) m++;
    if (brr[i] == -1) {
      brr[i] = m;
      fre[m] = 1;
    }
    cout << brr[i] << " ";
  }
  return 0;
}
