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;
  int arr[n];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  bool found = true;
  for (int i = 0; i < n; i++) {
    if (arr[i] > i + 1) {
      found = false;
      break;
    }
  }
  map<int, int> m;
  if (!found)
    cout << "-1" << endl;
  else {
    for (int i = 0; i < n; i++) m[arr[i]] = i;
    int b[n];
    for (int i = 0; i < n; i++) {
      b[i] = -1;
    }
    map<int, int>::iterator it;
    for (it = m.begin(); it != m.end(); it++) {
      if (it->second < n - 1) {
        b[it->second + 1] = it->first;
      }
    }
    int j = 0;
    for (int i = 0; i < n; i++) {
      if (b[i] != -1) continue;
      while (m.count(j) > 0) j++;
      b[i] = j;
      j++;
    }
    for (int i = 0; i < n; i++) cout << b[i] << " ";
    cout << endl;
  }
}
