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;
#pragma GCC optimize("Ofast")
template <typename T, typename U>
inline bool exist(T &cont, U &val) {
  return cont.find(val) != cont.end();
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int main(int argc, char **argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> v(n);
  set<int> s1;
  for (int i = 0, _n = (n); i < _n; i++) {
    cin >> v[i];
    s1.insert(v[i]);
    if (v[i] > i + 1) {
      cout << -1 << '\n';
      return 0;
    }
  }
  vector<int> ans(n, -1);
  vector<int> ranges;
  for (int i = (1), _b = (n - 1); i <= _b; i++) {
    if (v[i - 1] != v[i]) {
      ans[i] = v[i - 1];
    }
  }
  int start = !v[0];
  for (int i = 0, _n = (n); i < _n; i++) {
    if (ans[i] == -1) {
      while (s1.find(start) != s1.end()) {
        start++;
      }
      ans[i] = start++;
    }
  }
  for (int i = 0, _n = (n); i < _n; i++) {
    cout << ans[i] << ' ';
  }
  return 0;
}
