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;
const int N = 2e5 + 5, mod = 1e9 + 7;
int test = 1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  int n;
  cin >> n;
  int a[n + 1];
  int mex = 0, f = 1;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] > (i + 1)) f = 0;
  }
  int b[n + 1];
  if (f == 0) {
    cout << -1 << endl;
    exit(0);
  }
  map<int, int> m;
  int d = n;
  for (int i = n - 1; i >= 0; i--) {
    m[a[i]]++;
    if (a[i - 1] != a[i] && i != 0) {
      b[i] = a[i - 1];
      m[a[i - 1]]++;
    } else {
      while (m[d] > 0) d--;
      b[i] = d;
      d--;
    }
  }
  for (int i = 0; i < n; i++) cout << b[i] << " ";
  cout << endl;
}
