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 n;
int a[100007], b[100007], vis[100007];
void init() {
  for (int i = 0; i <= n; i++) {
    vis[i] = 0;
    b[i] = -1;
  }
}
int main() {
  cin >> n;
  int ok = 1;
  init();
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > i) ok = 0;
  }
  if (ok == 0)
    printf("-1\n");
  else {
    vis[a[n]] = 1;
    for (int i = 1; i <= n; i++) {
      if (a[i] != a[i - 1]) {
        b[i] = a[i - 1];
        vis[b[i]] = 1;
      }
    }
    int l = 0;
    for (int i = 1; i <= n; i++) {
      if (b[i] == -1) {
        while (vis[l]) {
          l++;
        }
        vis[l] = 1;
        b[i] = l;
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", b[i]);
    }
    puts("");
  }
}
