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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int arr[n], a[n], i;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    a[i] = arr[i];
  }
  int ans[n];
  map<int, int> done, mark;
  if (arr[0] == 0) {
    ans[0] = 1;
    done[1] = 1;
  } else {
    ans[0] = 0;
    done[0] = 1;
  }
  mark[0] = 1;
  int start = 0, flag = 0;
  for (i = 1; i < n; i++) {
    if (a[i] != a[i - 1]) {
      if (done[a[i - 1]] == 1) {
        flag = 1;
        break;
      }
      ans[i] = a[i - 1];
      done[a[i - 1]] = 1;
      mark[i] = 1;
    }
  }
  done[a[n - 1]] = 1;
  if (flag == 1) {
    cout << "-1" << endl;
  } else {
    for (i = 0; i < n; i++) {
      if (mark[i] == 1) {
        cout << ans[i] << " ";
      } else {
        while (done[start] != 0) {
          start++;
        }
        done[start] = 1;
        cout << start << " ";
      }
    }
  }
  return 0;
}
