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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int vis[100005];
long long int arr[100005];
long long int ans[100005];
int32_t main() {
  c_p_c();
  long long int n;
  cin >> n;
  long long int f = 0;
  for (long long int i = 1; i <= n; i++) {
    cin >> arr[i];
    if (arr[i] > i) {
      f = 1;
    }
    vis[arr[i]] = 1;
  }
  if (f) {
    cout << -1 << "\n";
    return 0;
  }
  memset(ans, -1, sizeof(ans));
  for (long long int i = 2; i <= n; i++) {
    if (arr[i] != arr[i - 1]) {
      ans[i] = arr[i - 1];
      vis[arr[i - 1]] = 1;
    }
  }
  vector<long long int> not_vis;
  for (long long int i = 0; i <= 100004; i++) {
    if (vis[i] != 1) {
      not_vis.push_back(i);
    }
  }
  long long int j = 0;
  for (long long int i = 1; i <= n; i++) {
    if (ans[i] == -1) {
      ans[i] = not_vis[j++];
    }
  }
  for (long long int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
  return 0;
}
