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;
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
  uniform_int_distribution<int> uid(0, lim - 1);
  return uid(rang);
}
long long INF = LLONG_MAX;
const long long M = 1000000007;
long long powm(long long, long long);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  map<long long, long long> mp;
  long long n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    mp[x]++;
  }
  for (auto x : mp) {
    long long fi = x.first;
    long long se = x.second;
    long long m = se / 2;
    mp[fi] %= 2;
    if (m) mp[fi + 1] += m;
  }
  long long ct = 0;
  for (auto x : mp)
    if (x.second) ct++;
  long long hi = 0;
  long long sz = mp.size();
  for (auto x : mp) hi = max(hi, x.first);
  cout << hi - ct + 1;
}
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % M;
    a = (a * a) % M;
    b >>= 1;
  }
  return res;
}
