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 main() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  set<long long> nums;
  for (long long i = 0; i <= n; i++) {
    nums.insert(i);
  }
  vector<long long> b(n);
  vector<long long> c;
  vector<long long> d;
  for (long long i = 0; i < n; i++) {
    d.push_back(i);
    c.clear();
    while (*nums.begin() < a[i]) {
      c.push_back(*nums.begin());
      nums.erase(nums.begin());
    }
    reverse(c.begin(), c.end());
    if (d.size() < c.size()) {
      cout << -1 << endl;
      return 0;
    }
    while (!c.empty()) {
      auto it = lower_bound(a.begin(), a.end(), c.back());
      if (it != a.end() && *it == c.back() &&
          distance(a.begin(), it) >= d.back()) {
        cout << -1 << endl;
        return 0;
      }
      b[d.back()] = c.back();
      c.pop_back();
      d.pop_back();
    }
  }
  for (long long x : d) {
    b[x] = n + 1;
  }
  for (long long i = 0; i < n; i++) {
    cout << b[i] << " ";
  }
  cout << endl;
  return 0;
}
