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;
int main() {
  int n;
  cin >> n;
  int v[n];
  bool possible = 1;
  bool tab[1000042] = {0};
  int valMax = 0;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    valMax = max(valMax, v[i]);
    if (v[i] > i + 1) {
      possible = 0;
    }
    tab[v[i]] = 1;
  }
  if (!possible) {
    cout << "-1";
    return 0;
  }
  int mex = 0;
  vector<int> manque;
  for (int i = 0; i <= valMax; i++) {
    if (!tab[i]) {
      manque.push_back(i);
    }
  }
  int curManque = 0;
  vector<int> rep;
  for (int i = 0; i < n; i++) {
    if (mex == v[i]) {
      if (curManque < manque.size() && manque[curManque] <= mex) {
        curManque++;
      }
      if (curManque >= manque.size()) {
        rep.push_back(valMax + 1);
        if (valMax == 1000000) {
          cout << "-1";
          return 0;
        }
      } else {
        rep.push_back(manque[curManque]);
        curManque++;
      }
    } else {
      rep.push_back(mex);
      mex = v[i];
    }
  }
  for (int i = 0; i < rep.size(); i++) cout << rep[i] << " ";
  return 0;
}
