Ivan has got an array of n non-negative integers a1, a2, ..., an. Ivan knows that the array is sorted in the non-decreasing order. 

Ivan wrote out integers 2a1, 2a2, ..., 2an on a piece of paper. Now he wonders, what minimum number of integers of form 2b (b ≥ 0) need to be added to the piece of paper so that the sum of all integers written on the paper equalled 2v - 1 for some integer v (v ≥ 0). 

Help Ivan, find the required quantity of numbers.

Input

The first line contains integer n (1 ≤ n ≤ 105). The second input line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 2·109). It is guaranteed that a1 ≤ a2 ≤ ... ≤ an.

Output

Print a single integer — the answer to the problem.

Examples

Input

4
0 1 1 1


Output

0


Input

1
3


Output

3

Note

In the first sample you do not need to add anything, the sum of numbers already equals 23 - 1 = 7.

In the second sample you need to add numbers 20, 21, 22.
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-ffloat-store")
#pragma GCC optimize("-fno-defer-pop")
long long int power(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  if (b == 1) return a % m;
  long long int t = power(a, b / 2, m) % m;
  t = (t * t) % m;
  if (b & 1) t = ((t % m) * (a % m)) % m;
  return t;
}
long long int modInverse(long long int a, long long int m) {
  return power(a, m - 2, m);
}
map<long long int, long long int> mask;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k, l, n;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> k;
    mask[k] += 1;
  }
  long long int c = 0, an = 0, b = 0;
  for (auto it : mask) {
    c = max(c, it.first);
    b = it.second / 2;
    it.second = it.second % 2;
    an += (it.second == 1);
    if (b >= 1) {
      mask[it.first + 1] += b;
    }
  }
  cout << c - an + 1;
  return 0;
}
