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, i, j = 0;
  cin >> n;
  int a[n];
  unordered_set<int> s, v;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    s.insert(a[i]);
  }
  vector<int> p;
  while (s.find(j) != s.end()) j++;
  v.insert(j);
  p.push_back(j++);
  for (i = 0; i < n; i++) {
    if (a[i] != 0 && v.find(a[i] - 1) == v.end()) break;
    if (i + 1 < n && a[i + 1] != a[i]) {
      v.insert(a[i]);
      p.push_back(a[i]);
    } else {
      while (s.find(j) != s.end()) j++;
      v.insert(j);
      p.push_back(j);
      j++;
    }
  }
  if (i < n)
    cout << -1 << endl;
  else {
    for (i = 0; i < n; i++) cout << p[i] << " ";
  }
}
