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>
inline long long int max(long long int a, long long int b) {
  return ((a > b) ? a : b);
}
inline long long int min(long long int a, long long int b) {
  return ((a > b) ? b : a);
}
inline long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
using namespace std;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long int modInverse(long long int a) { return power(a, 1000000007 - 2); }
const int N = 500023;
struct HASH {
  size_t operator()(const pair<int, int>& x) const {
    return hash<long long>()(((long long)x.first) ^
                             (((long long)x.second) << 32));
  }
};
bool compare(long long int x, long long int y) { return x > y; }
set<long long int> s;
void solve() {
  long long int n, a, k = 0;
  cin >> n;
  while (n--) {
    cin >> a;
    while (s.count(a)) {
      s.erase(a);
      a++;
    }
    s.insert(a);
  }
  cout << *s.rbegin() - s.size() + 1;
  cout << "\n";
  return;
}
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  };
  int T = 1;
  int t = 0;
  while (t++ < T) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}
