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 binary_search(int A[], int n, int x) {
  int left, right, mid;
  left = 0;
  right = n - 1;
  while (left <= right) {
    mid = (left + right) / 2;
    if (A[mid] == x) {
      return mid;
    }
    if (A[mid] < x) {
      left = mid + 1;
    } else {
      right = mid - 1;
    }
  }
  return -1;
}
int main() {
  int n;
  cin >> n;
  int ara[n], b[n], temp[n];
  for (int i = 0; i < n; i++) cin >> ara[i];
  int k = 0;
  for (int i = 1; i < n; i++)
    if (binary_search(ara, n, i) == -1) temp[k++] = i;
  if (ara[0] == 0)
    b[0] = temp[0];
  else
    b[0] = 0;
  cout << b[0] << " ";
  for (int i = 1, m = 1; i < n; i++) {
    if (ara[i] == ara[i - 1]) {
      if (m != k)
        b[i] = temp[m++];
      else
        b[i] = temp[m - 1];
    } else
      b[i] = ara[i - 1];
    cout << b[i] << " ";
  }
  cout << endl;
  return 0;
}
