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;
int main() {
  int n;
  cin >> n;
  vector<int> a(n), b(n, -1);
  for (int i = 0; i < n; i++) cin >> a[i];
  if (a[0] == 1)
    b[0] = 0;
  else
    b[0] = 1;
  for (int i = 1; i < n; i++) {
    if (a[i] != a[i - 1]) b[i] = a[i - 1];
  }
  int i = 1;
  bool ans = true;
  while (i < n) {
    if (b[i] != -1) {
      i++;
      continue;
    }
    {
      auto itr = upper_bound(a.begin(), a.end(), a[i]);
      int idx = itr - a.begin();
      b[i] = max(b[i - 1] + 1, a[i - 1] + 1);
      if (idx < n && b[i] >= a[idx]) b[i] = b[i - 1];
      b[i] = min(b[i], i + 1);
    }
    i++;
  }
  vector<int> temp = b;
  int mx = *max_element(b.begin(), b.end());
  vector<bool> p(mx + 1, false);
  for (auto cur : b) p[cur] = true;
  int c = 0;
  for (int i = 0; i <= mx; i++)
    if (!p[i]) c++;
  if (c <= 1) {
    for (i = 0; i < n; i++) cout << b[i] << " ";
  } else
    cout << "-1";
}
