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;
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& p) {
  os << "(" << p.first << "," << p.second << ") ";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto i : v) os << i << " ";
  return os;
}
const long long int mxN = 2e5;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n;
  cin >> n;
  set<long long int> s1;
  long long int a[n];
  for (long long int i = 0; i < n; i++) cin >> a[i], s1.insert(a[i]);
  set<long long int> s2;
  for (long long int i = 0; i <= n; i++) {
    if (s1.find(i) == s1.end()) s2.insert(i);
  }
  for (long long int i = 0; i < n; i++) {
    if (i && a[i] != a[i - 1]) s2.insert(a[i - 1]);
    cout << (*s2.begin()) << " ";
    s2.erase(s2.begin());
  }
}
