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;
int vis[1000006];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> vec, ans;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    vec.push_back(x);
    vis[x]++;
  }
  set<int> se;
  for (int i = 0; i <= 5 * n; i++) {
    se.insert(i);
  }
  for (auto x : vec) se.erase(x);
  for (int i = 0; i < n; i++) {
    if (vec[i] > i + 1) {
      return cout << -1 << "\n", 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << *se.begin() << " ";
    se.erase(se.begin());
    vis[vec[i]]--;
    if (vis[vec[i]] == 0) {
      se.insert(vec[i]);
    }
  }
  return 0;
}
