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;
using LL = long long;
LL arr[1000000], ans[1000000];
deque<int> pos;
map<int, int> ok;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  if (arr[1] == 0) {
    ans[0] = 1;
  }
  ok[ans[0]] = 1;
  int mex = 0;
  for (int i = 1; i <= n; i++) {
    if (ok[arr[i]] == 1) {
      cout << "-1" << endl;
      return 0;
    }
    if (arr[i] == mex) {
      pos.push_back(i);
      ans[i] = ans[i - 1];
    } else {
      int tmp = mex;
      pos.push_back(i);
      while (tmp < arr[i] && !pos.empty()) {
        ans[pos.back()] = tmp;
        pos.pop_back();
        ok[tmp] = 1;
        tmp++;
      }
      if (tmp != arr[i]) {
        cout << "-1" << endl;
        return 0;
      }
      mex = arr[i];
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}
