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 = 2e6 + 1000;
long long int tt, n, m, k, a[N], b[N];
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  bool bb = false;
  vector<long long int> v;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] > i + 1) bb = true;
    if (!i) {
      v.push_back(a[i]);
    } else if (a[i] > a[i - 1]) {
      v.push_back(a[i]);
    }
  }
  if (bb) return cout << -1, 0;
  v.push_back(10000000);
  long long int pt = 0, h = 0;
  for (long long int i = 0; i < n; i++) {
    while (h == v[pt]) {
      h++;
      pt++;
    }
    if (i && a[i] > a[i - 1]) {
      cout << a[i - 1] << " ";
    } else {
      cout << h << " ";
    }
    h = min(h + 1, n + 100);
  }
  return 0;
}
