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 double EPS = 1e-9, PI = acosl(-1.);
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7, MOD2 = 2e9 + 11;
const int N = 1e5 + 5;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  int zero = -1;
  vector<int> v(n + 1);
  vector<int> ans(n + 1);
  vector<bool> visited(n + 1, false);
  v[0] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    if (v[i] > v[i - 1]) {
      ans[i] = v[i - 1];
      visited[v[i - 1]] = true;
      if (!ans[i]) zero = i;
    } else
      ans[i] = 0;
  }
  visited[v[n]] = true;
  int cont = 0;
  for (int i = 1; i <= n; i++) {
    while (visited[cont]) cont++;
    if (!ans[i] && i != zero) ans[i] = cont, visited[cont] = true;
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}
