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;
const long long int mod = 1e9 + 7;
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> arr(n);
  for (long long int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  vector<long long int> ans(n);
  long long int j = 0;
  long long int k = 0;
  if (arr[0] == 0)
    ans[0] = 1;
  else
    ans[0] = 0;
  for (long long int i = 1; i < n; ++i) {
    if (arr[i] != arr[i - 1]) {
      ans[i] = arr[i - 1];
    } else {
      ans[i] = arr[i] + 1;
    }
  }
  for (long long int i = 0; i < n; ++i) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int tc = 1;
  for (long long int i = 1; i <= tc; ++i) {
    solve();
  }
  return 0;
}
