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;
  int arr[100001];
  scanf("%d", &n);
  bool valid = true;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] > i + 1) {
      printf("-1\n");
      return 0;
    }
  }
  int sol[100001];
  int cur = 0;
  int ptr = -1;
  int used = 0;
  for (int i = 0; i < n; i++) {
    if (arr[i] > cur) {
      int num = cur + 1;
      int next = -1;
      for (int j = ptr + 1; j < i; j++) {
        if (sol[j] != 100001) {
          used--;
          continue;
        }
        if (num == arr[i]) {
          next = j - 1;
          break;
        }
        sol[j] = num++;
      }
      if (num != arr[i]) {
        printf("-1\n");
        return 0;
      }
      if (next == -1) {
        next = i - 1;
      }
      ptr = next;
      sol[i] = cur;
      cur = arr[i];
      used++;
    } else {
      sol[i] = 100001;
    }
  }
  if (!valid) {
    printf("-1\n");
    return 0;
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", sol[i]);
  }
  printf("\n");
  return 0;
}
