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 ans = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return ans * f;
}
const int N = 1e5 + 5;
int n, a[N], b[N], have, num, now = 1;
int main() {
  memset(b, -1, sizeof(b));
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i <= n; ++i) {
    if (a[i] > i) {
      printf("-1\n");
      return 0;
    }
    for (; num <= a[i] - 1; ++num) {
      if (b[i] == -1)
        b[i] = num;
      else {
        while (b[now] != -1) now++;
        b[now] = num;
      }
    }
    num = a[i];
  }
  for (int i = 1; i <= n; ++i)
    if (b[i] == -1)
      printf("%d ", 1000000);
    else
      printf("%d ", b[i]);
  return 0;
}
