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;
void solve() {
  int n;
  cin >> n;
  int a[n + 1];
  vector<int> b(n);
  map<int, int> mp;
  for (int i = 1; i <= n; i++) cin >> a[i], mp[a[i]] = i;
  for (int i = 1; i <= n; i++)
    if (a[i] > i) {
      cout << "-1";
      return;
    }
  int k = 0;
  deque<int> q;
  for (auto i : mp) q.push_back(i.first);
  for (int i = 1; i <= n; i++) {
    if (a[i] == q.front()) {
      while (mp.find(k) != mp.end()) k++;
      b[i - 1] = k++;
    } else {
      mp.erase(q.front());
      b[i - 1] = q.front();
      q.pop_front();
    }
  }
  for (int i : b) cout << i << " ";
  cout << endl;
}
int main() {
  int T;
  T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
