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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
int main() {
  long long i, j, n;
  cin >> n;
  vector<long long> a(n);
  for (i = 0; i < n; i++) cin >> a[i];
  set<long long> st;
  for (auto in : a) st.insert(in);
  vector<long long> v1;
  for (auto in : st) v1.push_back(in);
  map<long long, long long> m;
  for (auto in : st) m[in] = 1;
  vector<long long> v2;
  for (i = 0; i <= 1000000; i++)
    if (m[i] == 0) v2.push_back(i);
  long long p1 = 0, p2 = 0;
  vector<long long> ans;
  for (i = 0; i < n; i++) {
    if (i == 0 || a[i] == a[i - 1])
      ans.push_back(v2[p2]), p2++;
    else
      ans.push_back(v1[p1]), p1++;
  }
  for (auto in : ans) cout << in << " ";
  cout << endl;
}
