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;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long tests = 1LL;
  for (int testno = 1; testno < tests + 1; testno++) {
    long long n;
    cin >> n;
    long long a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    long long ans[n];
    for (int i = 0; i < n; i++) ans[i] = -1;
    if (a[0] == 0)
      ans[0] = 1000000;
    else
      ans[0] = a[0] - 1;
    long long lst = 0, f = 1;
    for (int i = 1; i < n; i++) {
      if (i == 0 or a[i] == a[i - 1]) {
        continue;
      } else {
        ans[i] = a[i - 1];
        long long df = a[i] - a[i - 1] - 1;
        long long k = i - lst - 1;
        if (df == 0 and k != 0) {
          f = 0;
          break;
        } else {
          for (int h = i - 1; h > lst; h--) {
            ans[h] = min(a[i] - 1, ans[h + 1] + 1);
          }
          lst = i;
        }
      }
    }
    if (f == 1) {
      for (int i = 0; i < n; i++) cout << ans[i] << " ";
      cout << '\n';
    } else
      cout << "-1\n";
  }
  return 0;
}
