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, k, x, y;
  cin >> n;
  x = 0;
  int h[n + 3], a[n];
  fill(h, h + n + 3, 0);
  for (i = 0; i < n; i++) {
    cin >> k;
    a[i] = k;
    h[k] = 1;
  }
  int b[n];
  x = 0;
  y = 0;
  for (i = 0; i < n; i++) {
    if (i == 0) {
      b[i] = (a[i]) ^ (1);
      h[a[i] ^ 1] = 1;
    } else
      b[i] = a[i - 1];
    for (j = i + 1; j < n; j++) {
      if (a[j] != a[i]) break;
      while (h[y] == 1) {
        y++;
      }
      b[j] = y;
      h[y] = 1;
    }
    i = j - 1;
  }
  for (i = 0; i < n; i++) cout << b[i] << " ";
  cout << '\n';
}
