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;
template <typename T>
void printvec(vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++) {
    cout << vec[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long int n;
  cin >> n;
  vector<long long int> inp(n);
  for (int i = 0; i < n; i++) {
    cin >> inp[i];
  }
  vector<long long int> ans(n, -1);
  vector<long long int> sp;
  long long int mex = 0;
  for (int i = 0; i < n; i++) {
    if (i != 0) {
      if (inp[i] < inp[i - 1]) {
        cout << "-1\n";
        return 0;
      }
    }
    if (mex != inp[i]) {
      ans[i] = mex;
      mex++;
      while (mex != inp[i]) {
        if (sp.size() == 0) {
          cout << "-1\n";
          return 0;
        }
        int ind = sp.back();
        sp.pop_back();
        ans[ind] = mex;
        mex++;
      }
      continue;
    } else if (mex == inp[i]) {
      sp.push_back(i);
    }
  }
  while (sp.size() != 0) {
    int ind = sp.back();
    sp.pop_back();
    ans[ind] = mex + 1;
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
}
