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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int64_t n;
  cin >> n;
  vector<int64_t> v(n);
  map<int64_t, int64_t> present;
  for (int64_t i = 0; i < n; i++) {
    cin >> v[i];
    present[v[i]]++;
  }
  set<int64_t> s;
  vector<int64_t> ans;
  bool flag = 0;
  set<int64_t> left;
  int64_t startWith = 0;
  for (int64_t i = 0; i < n; i++) {
    if (s.count(v[i])) {
      flag = 1;
      break;
    }
    for (set<int64_t>::iterator it = left.begin(); it != left.end();) {
      if ((*it) >= v[i]) {
        break;
      }
      s.insert(*it);
      ans.push_back(*it);
      set<int64_t>::iterator it2 = it;
      it++;
      left.erase(it2);
    }
    if (ans.size() == i + 1) {
      present[v[i]]--;
      left.insert(v[i]);
      continue;
    }
    if (ans.size() > i + 1) {
      flag = 1;
      break;
    }
    startWith = max(v[i] + 1, startWith);
    if (startWith == v[i]) {
      left.insert(startWith);
      ++startWith;
    }
    while (present[startWith] > 0) {
      left.insert(startWith);
      ++startWith;
    }
    ans.push_back(startWith);
    s.insert(startWith);
    ++startWith;
    present[v[i]]--;
    left.insert(v[i]);
  }
  if (flag) {
    cout << -1 << endl;
    return 0;
  }
  for (int64_t j = ans[ans.size() - 1]; ans.size() < n;) {
    ans.push_back(j);
  }
  for (int64_t i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}
