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 N = 1e6 + 10;
const long long int mod = 1e9 + 7;
const double pi = acos(-1);
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int n;
  cin >> n;
  vector<long long int> a(n), have(N, 0);
  for (long long int i = 0; i < n; ++i) {
    cin >> a[i];
    have[a[i]]++;
  }
  long long int mex = 0;
  vector<long long int> ans(N, -1);
  for (long long int i = 0; i < n; ++i) {
    if (ans[i] == -1) {
      while (have[mex] > 0) ++mex;
      ans[i] = mex;
      ++mex;
    }
    --have[a[i]];
    if (have[a[i]] == 0) {
      ans[i + 1] = a[i];
      mex = max(mex, a[i] + 1);
    }
  }
  for (long long int i = 0; i < n; ++i) cout << ans[i] << ' ';
  cout << '\n';
  return 0;
}
