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;
  int *a = new int[n];
  vector<int> not_req;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i != 0 && a[i] != a[i - 1]) {
      int count = a[i] - a[i - 1] - 1;
      for (int j = 1; j <= count; j++) {
        not_req.push_back(a[i - 1] + j);
      }
    }
  }
  if (a[0] > 1) {
    cout << -1 << endl;
    delete[] a;
    return 0;
  }
  int *b = new int[n];
  if (a[0] == 1) {
    b[0] = 0;
  } else {
    b[0] = 1;
  }
  int k = not_req.size();
  int j = 0;
  bool flag = true;
  for (int i = 1; i < n; i++) {
    if (a[i] == a[i - 1]) {
      if (j < k && i + 1 < n && not_req[j] < a[i + 1]) {
        b[i] = not_req[j];
        j++;
      } else {
        b[i] = b[i - 1];
      }
    } else {
      b[i] = a[i - 1];
      if (j == k || not_req[j] > a[i])
        continue;
      else {
        flag = false;
        break;
      }
    }
  }
  if (flag) {
    for (int i = 0; i < n; i++) cout << b[i] << " ";
    cout << endl;
  } else
    cout << -1 << endl;
  delete[] a;
  delete[] b;
}
