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;
long long int fastMax(long long int x, long long int y) {
  return (((y - x) >> (32 - 1)) & (x ^ y)) ^ y;
}
long long int fastMin(long long int x, long long int y) {
  return (((y - x) >> (32 - 1)) & (x ^ y)) ^ x;
}
const long long int MAXN = 1e5 + 10;
long long int a[MAXN], b[MAXN];
deque<long long int> empty_pos;
void solve() {
  long long int n;
  cin >> n;
  for (long long int i = (1); i <= (n); ++i) {
    cin >> a[i];
  }
  long long int val = 0;
  for (long long int i = (1); i <= (n); ++i) {
    if (a[i] == val)
      empty_pos.push_back(i);
    else {
      empty_pos.push_back(i);
      long long int n_val = a[i] - 1, idx = i;
      while (n_val > val) {
        b[empty_pos.front()] = n_val--;
        empty_pos.pop_front();
      }
      b[empty_pos.back()] = n_val--;
      empty_pos.pop_back();
      val = a[i];
    }
  }
  while ((long long int)empty_pos.size()) {
    b[empty_pos.front()] = n + 1;
    empty_pos.pop_front();
  }
  for (long long int i = (1); i <= (n); ++i) cout << b[i] << " ";
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  for (long long int i = (1); i <= (t); ++i) {
    solve();
  }
}
