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;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long solve(long long st, long long ed, long long sum, long long x,
                long long arr[]) {
  if (ed < st) return -1;
  if (sum % x != 0) return ed - st + 1;
  return max(solve(st, ed - 1, sum - arr[ed], x, arr),
             solve(st + 1, ed, sum - arr[st], x, arr));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    for (long long i = 0; i < n; i++) cin >> arr[i];
    ;
    long long ans[n];
    long long b = 0;
    set<long long> sa;
    set<long long> sb;
    sb.insert(0);
    long long prev = b;
    for (long long i = 0; i < n; i++) sa.insert(arr[i]);
    for (long long i = 0; i < n; i++) {
      sa.insert(arr[i]);
      long long temp = b;
      while (sa.find(temp) != sa.end()) temp++;
      prev = temp;
      ans[i] = temp;
      sa.insert(temp);
      if (temp == b) {
        if (i >= 1)
          b += ans[i - 1] + 1;
        else
          b++;
      }
      sa.erase(sa.find(arr[i]));
    }
    for (long long i = 0; i < n; i++) cout << ans[i] << " ";
    ;
    cout << "\n";
  }
  return 0;
}
