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 mxn = 100005;
long long int arr[mxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[100005], b[100005] = {}, i, x;
    long long int c[100005] = {};
    for (i = 0; i < n; i++) {
      cin >> a[i];
      c[a[i]]++;
    }
    vector<int> v;
    int flag = 0;
    queue<int> ans;
    for (i = 0; i <= n; i++) {
      if (c[i] == 0) continue;
      while (c[i]) {
        if (ans.size()) {
          c[i]--;
          v.push_back(ans.front());
          b[ans.front()]++;
          ans.pop();
          continue;
        }
        while (c[flag] or b[flag]) flag++;
        b[flag]++;
        v.push_back(flag++);
        c[i]--;
      }
      ans.push(i);
    }
    for (i = 0; i < n; i++) cout << v[i] << " ";
    cout << endl;
  }
  return 0;
}
