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;
struct type {
  long long k, i, j;
};
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long findSubarray(long long arr[], long long n, long long k) {
  long long window_sum = 0;
  long long min_window = INT_MAX;
  long long last = 0;
  for (long long i = 0; i < n; i++) {
    window_sum += arr[i];
    if (i + 1 >= k) {
      if (min_window > window_sum) {
        min_window = window_sum;
        last = i;
      }
      window_sum -= arr[i + 1 - k];
    }
  }
  long long sum = 0;
  for (long long i = last - k + 1; i <= last; i++) {
    sum += arr[i];
  }
  return sum;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  vector<long long> a(n), b;
  for (long long i = 0; i < n; i++) cin >> a[i];
  set<long long> s;
  for (long long i = 0; i < n; i++) s.insert(a[i]);
  vector<long long> c;
  for (long long i = 0; i <= n; i++) {
    if (s.find(i) == s.end()) {
      c.push_back(i);
    }
  }
  b.push_back(c[0]);
  long long x = 1;
  for (long long i = 1; i < n; i++) {
    if (a[i] == a[i - 1]) {
      b.push_back(c[x]);
      x++;
    } else {
      b.push_back(a[i - 1]);
    }
  }
  for (auto x : b) cout << x << " ";
  cout << "\n";
}
