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 nextPow(long long n) {
  long long p = 1;
  while (p <= n) {
    p *= 2;
  }
  return p;
}
int main() {
  ios_base::sync_with_stdio(0);
  unordered_map<long long, long long> cnt;
  int n;
  cin >> n;
  int a[n];
  vector<long long> u;
  long long c = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i > 0 && a[i] != a[i - 1]) {
      cnt[a[i - 1]] = c;
      c = 1;
      u.push_back(a[i - 1]);
    } else {
      c++;
    }
  }
  cnt[a[n - 1]] = c;
  u.push_back(a[n - 1]);
  long long ans = 0;
  long long prev = -1;
  queue<long long> q;
  for (int i = 0; i < u.size() || !q.empty();) {
    if (q.empty()) {
      ans += u[i] - 1 - prev;
      q.push(u[i++]);
    }
    long long top = q.front();
    q.pop();
    if (i < u.size() && top == u[i]) {
      i++;
    }
    c = cnt[top];
    if (c > 1) {
      c--;
      int next = c;
      if (c % 2 == 1) {
        ans++;
        next++;
      }
      cnt[top + 1] += next / 2;
      q.push(top + 1);
    }
    prev = top;
  }
  cout << ans;
  return 0;
}
