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 t, n, a[100010], b[100010];
int f[1000010], exist[1000010];
int main() {
  for (int i = 0; i < 100010; ++i) {
    b[i] = -1;
  }
  scanf("%d ", &n);
  for (int j = 1; j <= n; ++j) {
    scanf("%d", &a[j]);
    exist[a[j]] = j;
    if (a[j] > j) {
      puts("-1");
      return 0;
    }
    if (j >= 2 && a[j - 1] < a[j]) {
      b[j] = a[j - 1];
      f[b[j]] = 1;
    }
  }
  int num = 0;
  for (int i = 1; i <= n; ++i) {
    if (b[i] != -1) continue;
    while (f[num] != 0) {
      num++;
    }
    if (i <= exist[num]) {
      b[i] = num + 1;
    } else {
      b[i] = num;
    }
    f[num] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d ", b[i]);
  }
  return 0;
}
