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;
const long long N = 2 * 1e5 + 4;
long long add(long long a, long long b, long long m) {
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = ((a % m) + (b % m)) % m;
  res %= m;
  return res;
}
long long mul(long long a, long long b, long long m) {
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = ((a % m) * (b % m)) % m;
  res %= m;
  return res;
}
long long sub(long long a, long long b, long long m) {
  long long res = (a - b + m) % m;
  res %= m;
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> res(n, -1);
  set<long long> second;
  long long prev = 0;
  for (long long i = 0; i < n; i++) {
    long long cur;
    cin >> cur;
    second.insert(cur);
    if (cur != prev) res[i] = prev;
    prev = cur;
  }
  second.insert(0);
  long long idx = 0;
  for (long long i = 0; i < n; i++) {
    if (res[i] != -1) continue;
    while (second.find(idx) != second.end()) {
      idx++;
      cerr << idx << '\n';
    }
    second.insert(idx);
    res[i] = idx;
  }
  for (auto x : res) cout << x << " ";
  return 0;
}
