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;
long long int modu(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % 1000000007;
    b /= 2;
    a = (a * a) % 1000000007;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  long long int arr[n + 1];
  set<long long int> s;
  for (long long int i = 1; i <= n; i++) {
    cin >> arr[i];
    s.insert(arr[i]);
  }
  long long int cn = 0;
  vector<long long int> ans;
  while (s.find(cn) != s.end()) cn++;
  ans.push_back(cn);
  cn++;
  long long int i = 2;
  while (i <= n) {
    if (arr[i] == arr[i - 1]) {
      while (s.find(cn) != s.end()) cn++;
      ans.push_back(cn);
      cn++;
      i++;
    } else {
      ans.push_back(arr[i - 1]);
      i++;
    }
  }
  for (long long int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
