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 a[100010];
int b[100010];
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  queue<int> empty_location;
  int k = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] > k) {
      b[i] = k;
      for (int j = a[i] - 1; j > k; j--) {
        if (empty_location.empty()) {
          cout << "-1" << endl;
          return 0;
        }
        int location = empty_location.front();
        empty_location.pop();
        b[location] = j;
      }
      k = a[i];
    } else {
      empty_location.push(i);
    }
  }
  while (!empty_location.empty()) {
    int location = empty_location.front();
    empty_location.pop();
    b[location] = k + 1;
  }
  for (int i = 0; i < n; i++) {
    cout << b[i] << endl;
  }
  return 0;
}
