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 n;
set<long long> s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  vector<long long> v(n), ans(n);
  for (int i = 0; i <= 2 * n; i++) s.insert(i);
  bool ok = 1;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] > i + 1) ok = 0;
    if (s.find(v[i]) != s.end()) s.erase(v[i]);
  }
  if (!ok) {
    cout << -1;
    return 0;
  }
  if (v[0] == 0) {
    ans[0] = 1;
    if (s.find(1) != s.end()) s.erase(1);
  } else {
    ans[0] = 0;
    if (s.find(0) != s.end()) s.erase(0);
  }
  for (int i = 1; i < n; i++) {
    if (v[i] != v[i - 1])
      ans[i] = v[i - 1];
    else {
      ans[i] = *s.begin();
      s.erase(s.begin());
    }
  }
  for (long long& i : ans) cout << i << " ";
  return 0;
}
