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 double pi = 3.14159265358979;
const int mod = 1e9 + 7;
const int N = 1e6 + 7;
void solve() {
  int n;
  cin >> n;
  int a[n + 1];
  int ok = 1;
  map<int, int> mp;
  for (int i = 1; i < n + 1; i++) {
    cin >> a[i];
    if (a[i] > i) ok = 0;
    mp[a[i]] = 1;
  }
  if (!ok) {
    cout << -1;
    return;
  }
  int cur = 0;
  int ans[n + 1];
  for (int i = 1; i < n + 1; i++) {
    while (mp[cur]) cur++;
    if (i == 1) {
      ans[i] = cur++;
    } else {
      if (a[i] == a[i - 1]) {
        ans[i] = cur++;
      } else {
        ans[i] = a[i - 1];
      }
    }
    cout << ans[i] << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  while (tc--) {
    solve();
    cout << '\n';
  }
  return 0;
}
