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>
int a[101000];
int n;
int cnt[101000];
int ans[101000];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < a[n - 1]; i++) cnt[i] = 1;
  int last = a[n - 1] - 1;
  int cst = a[n - 1] + 1;
  for (int i = n - 1; i >= 0; i--) {
    bool flag = false;
    if (i == 0) {
      flag = (a[i] == 1) ? true : false;
      ans[i] = 0;
    } else {
      flag = (a[i] != a[i - 1]) ? true : false;
      ans[i] = a[i - 1];
    }
    if (!flag) {
      if (last > a[i]) {
        ans[i] = last;
        cnt[last] = 0;
      } else
        ans[i] = cst;
    }
    cnt[ans[i]] = 0;
    while (last >= 0 && !cnt[last]) last--;
  }
  for (int i = 0; i < a[n - 1]; i++)
    if (cnt[i] > 0) {
      printf("%d\n", -1);
      return 0;
    }
  for (int i = 0; i < n; i++) printf("%d ", ans[i]);
  return 0;
}
