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;
set<pair<long long, int> > st;
map<long long, long long> mp;
const int maxn = 100000 + 100;
long long n, a[maxn], cnt = 0, mx;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  a[n + 1] = -1;
  for (int s = 1, e = 1; s <= n;) {
    while (a[e] == a[s]) e++;
    mp[a[s]] = e - s;
    st.insert(make_pair(a[s], e - s));
    s = e;
  }
  long long x, y;
  while (st.size()) {
    x = st.begin()->first;
    y = st.begin()->second;
    if (y % 2 == 1) cnt++;
    int num = 0;
    y /= 2;
    while (y > 0) {
      num++;
      if (y % 2 == 1) {
        if (mp.find(x + num) == mp.end()) {
          mp[x + num] = 1;
          st.insert(make_pair(x + num, 1));
        } else {
          st.erase(make_pair(x + num, mp[x + num]));
          mp[x + num]++;
          st.insert(make_pair(x + num, mp[x + num]));
        }
      }
      y /= 2;
    }
    st.erase(st.begin());
  }
  cout << x - cnt + 1 << endl;
  return 0;
}
