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;
long long in() {
  long long a;
  scanf("%lld", &a);
  return a;
}
double din() {
  double a;
  scanf("%lf", &a);
  return a;
}
long long bigmod(long long b, long long p, long long md) {
  if (p == 0) return 1;
  if (p % 2 == 1) {
    return ((b % md) * bigmod(b, p - 1, md)) % md;
  } else {
    long long y = bigmod(b, p / 2, md);
    return (y * y) % md;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  int n = in();
  priority_queue<long long, vector<long long>, greater<long long> > pq;
  for (int i = 0; i < n; i++) {
    long long a = in();
    pq.push(a);
  }
  set<long long> s;
  long long mx = 0;
  while (!pq.empty()) {
    if (pq.size() == 1) {
      long long a = pq.top();
      mx = a;
      pq.pop();
      s.insert(a);
    } else {
      long long a = pq.top();
      pq.pop();
      long long b = pq.top();
      pq.pop();
      if (a == b) {
        pq.push(b + 1);
      } else {
        s.insert(a);
        pq.push(b);
      }
    }
  }
  printf("%lld\n", mx - s.size() + 1);
}
