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;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  std::vector<long long> b, s;
  long long a[n + 1], flag = 1, cur = 0;
  a[0] = -1;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] < a[i - 1]) flag = 0;
    if (a[i] - i > i - 1) flag = 0;
  }
  if (flag == 1) {
    for (long long i = 1; i <= n; i++) {
      for (long long j = a[i - 1]; j < a[i]; j++) {
        if (j != a[i - 1] && j != a[i]) s.push_back(j);
      }
    }
  }
  s.push_back(1e6);
  if (flag == 1) {
    b.push_back(s[cur]);
    cur++;
    for (long long i = 2; i <= n; i++) {
      if (a[i] == a[i - 1]) {
        b.push_back(s[cur]);
        if (cur < s.size() - 1) cur++;
      } else if (a[i] != a[i - 1]) {
        b.push_back(a[i - 1]);
      }
    }
    for (long long i = 0; i < b.size(); i++) cout << b[i] << ' ';
  } else
    cout << -1;
  return (0);
}
