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 main() {
  int n;
  cin >> n;
  vector<int> a(n), b(n, 1), cant(1e5 + 7, 0);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    cant[a[i]] += 1;
  }
  priority_queue<int, vector<int>, greater<int>> q;
  for (int i = 0; i < 1e5 + 7; ++i) {
    if (!cant[i]) q.push(i);
  }
  for (int i = 0, mx = 0; i < n; ++i) {
    b[i] = q.top();
    q.pop();
    cant[a[i]] -= 1;
    if (!cant[a[i]]) q.push(a[i]);
  }
  for (int& i : b) {
    cout << i << ' ';
  }
  return 0;
}
