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 int INF = 10e7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, i, f = 0, x;
  cin >> n;
  vector<long long int> v(n + 1), ans;
  set<long long int> s1, s2;
  for (i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] > i + 1) f = 1;
    x = v[i];
    s1.insert(x);
  }
  v[n] = INF;
  if (f == 1) {
    cout << "-1\n";
    return 0;
  }
  long long int mex = 0;
  for (i = 0; i < n; i++) {
    while (s1.find(mex) != s1.end()) mex++;
    if (s2.empty() == true) {
      ans.push_back(mex);
    } else {
      auto it = s2.begin();
      ans.push_back(*it);
      s2.erase(it);
    }
    if (v[i] != v[i + 1]) {
      auto it = s1.find(v[i]);
      s2.insert(*it);
    }
    if (s2.empty() == true) mex++;
  }
  for (auto it : ans) cout << it << " ";
  return 0;
}
