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>
long long mod = 1e9 + 7;
inline long long pwr(long long base, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n % 2 == 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    n /= 2;
  }
  return ans;
}
struct range {
  int l, h;
};
struct ab {
  int time, type;
};
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  map<int, int> m;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]] = 1;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] > i + 1) {
      cout << "-1\n";
      return 0;
    }
  }
  vector<int> ans;
  int c = 0;
  for (int i = 0; i < n; i++) {
    if (i == 0 || a[i] == a[i - 1]) {
      if (m[c]) {
        c++;
        i--;
      } else {
        ans.push_back(c);
        c++;
      }
    } else
      ans.push_back(a[i - 1]);
  }
  for (auto i : ans) cout << i << " ";
  return 0;
}
