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;
class Solution {
 public:
  vector<int> solve(vector<int>& A) {
    int n = A.size();
    vector<int> res(n, -1);
    deque<int> frees;
    int last = 0;
    for (int i = 0; i < n; ++i) {
      int cur = A[i];
      if (cur != last) {
        res[i] = last;
      } else {
        frees.push_back(i);
      }
      for (int j = last + 1; j < cur; ++j) {
        if (frees.empty()) {
          return {};
        }
        int p = frees.front();
        res[p] = j;
        frees.pop_front();
      }
      last = cur;
    }
    for (int i = 0; i < n; ++i) {
      if (res[i] == -1) {
        res[i] = ++last;
      }
    }
    return res;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution sol;
  int n;
  cin >> n;
  vector<int> A(n);
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
  }
  auto res = sol.solve(A);
  if (res.size() == n) {
    for (int i = 0; i < n; ++i) {
      cout << res[i] << (i == n - 1 ? '\n' : ' ');
    }
  } else {
    cout << -1 << '\n';
  }
  return 0;
}
