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;
void dbg_out() { cerr << "\b\b]\n"; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << H << ", ";
  dbg_out(T...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  map<long long, int> cnts;
  for (int i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    cnts[x]++;
  }
  long long ans = 0, curr = 0, cnt = 0;
  for (auto i : cnts) {
    long long a = i.first, b = i.second;
    if (curr == a) {
      if (cnt & 1) cnt++, ans++;
      cnt = b - 1 + cnt / 2;
      curr++;
      continue;
    }
    while (curr < a and cnt) {
      if (cnt & 1) {
        ans++;
        cnt++;
      }
      cnt /= 2;
      cnt--;
      curr++;
    }
    if (curr < a) {
      ans += (a - curr);
      curr = a;
    }
    if (cnt & 1) cnt++, ans++;
    cnt = b - 1 + cnt / 2;
    curr++;
  }
  while (cnt) {
    if (cnt & 1) {
      ans++;
      cnt++;
    }
    cnt /= 2;
    cnt--;
  }
  cout << ans;
  return 0;
}
