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 a[100010];
int vis[100010];
int b[100010];
int main() {
  int n;
  scanf("%d", &n);
  int flag = 1;
  for (register int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] > i) {
      flag = 0;
    }
  }
  if (flag == 0) {
    printf("-1\n");
  } else {
    memset(vis, 0, sizeof(vis));
    for (register int i = 1; i <= n; i++) {
      vis[a[i]] = 1;
    }
    int k = 0, cnt = 0;
    while (vis[k] == 1) {
      k++;
    }
    b[cnt++] = k;
    for (register int i = 2; i <= n; i++) {
      if (a[i] == a[i - 1]) {
        b[cnt++] = k;
      } else {
        b[cnt++] = a[i - 1];
        vis[a[i - 1]] = 1;
        while (vis[k] == 1) {
          k++;
        }
      }
    }
    for (register int i = 0; i < cnt; i++) {
      printf("%d ", b[i]);
    }
    printf("\n");
  }
  return 0;
}
