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 long pw(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans *= a;
    a *= a;
    b /= 2;
  }
  return (ans);
}
const long long mxn = 1e5 + 1;
long long a[mxn];
long long b[mxn];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    b[a[i]] = 1;
  }
  vector<long long> ans(n + 1);
  ans[0] = 1 - a[0];
  long long c = a[0];
  for (long long i = 1; i < n; i++) {
    if (a[i] != a[i - 1])
      ans[i] = a[i - 1];
    else {
      c = ans[i - 1] + 1;
      while (b[c]) c++;
      ans[i] = c;
      c++;
    }
  }
  for (long long i = 0; i < n; i++) cout << ans[i] << " ";
}
