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() {
  long long t;
  t = 1;
  long long i, j;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n];
    for (i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<long long> v, ans;
    if (a[0] == 0) {
      ans.push_back(1);
    } else {
      ans.push_back(0);
    }
    long long prev = a[0];
    i = 0;
    while (i < n) {
      while (i < n && a[i] == prev) {
        i++;
      }
      if (i < n) {
        for (j = prev + 1; j < a[i]; j++) {
          v.push_back(j);
        }
        prev = a[i];
      }
    }
    i = 1;
    prev = a[0];
    long long k = 0;
    while (i < n) {
      while (i < n && a[i] == prev) {
        i++;
        if (k < v.size()) {
          ans.push_back(v[k]);
          k++;
        } else {
          ans.push_back(v[k - 1]);
        }
      }
      if (i < n) {
        v.push_back(prev);
        prev = a[i];
      }
    }
    for (i = 0; i < ans.size(); i++) cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}
