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 Read() {
  int val = 0, opt = 1;
  char ch;
  while (!isdigit(ch = getchar()))
    if (ch == '-') opt = -1;
  while (isdigit(ch)) (val *= 10) += ch - '0', ch = getchar();
  return val * opt;
}
int a[100100], n, Vis[100100];
void work() {
  n = Read();
  for (int i = 1; i <= n; i++) {
    a[i] = Read();
    if (a[i] > i || a[i] < a[i - 1]) {
      printf("-1\n");
      return;
    }
    Vis[a[i]] = 1;
  }
  int t = 0;
  while (Vis[t]) t++;
  for (int i = 1; i <= n; i++) {
    if (a[i] != a[i - 1]) {
      printf("%d ", a[i - 1]);
      Vis[a[i - 1]] = 1;
    } else {
      while (Vis[t]) t++;
      printf("%d ", t);
      t++;
      while (Vis[t]) t++;
    }
  }
}
int main(void) {
  work();
  return 0;
}
