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>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:666000000")
using namespace std;
const int INF = (1 << 30) - 1;
const long double EPS = 1e-9;
void ML(const bool v) {
  if (v) return;
  int *ass;
  for (;;) {
    ass = new int[2500000];
    for (int i = 0; i < 2500000; i++) ass[i] = rand();
  }
}
void TL(const bool v) {
  if (v) return;
  for (;;) cout << rand() % (rand() % 1000 + 1) << endl;
}
void PE(const bool v) {
  if (v) return;
  for (int i = 0; i < 10000; i++) printf("%c", rand() % 256);
  exit(0);
}
int n;
vector<long long> a;
bool LoAd() {
  if (!(1 == scanf("%d", &n))) return false;
  a.resize(n);
  int u;
  for (int i = 0; i < (int)(n); i++) {
    (1 == scanf("%d", &u));
    a[i] = u;
  }
  return true;
}
void SoLvE() {
  sort(a.begin(), a.end());
  reverse(a.begin(), a.end());
  vector<long long> res;
  for (int i = 0; i < (int)(n); i++) {
    long long cur = a[i];
    while (!res.empty() && res.back() == cur) {
      res.pop_back();
      cur++;
    }
    res.push_back(cur);
  }
  long long ans = 0;
  for (int i = (int)res.size() - 1; i > 0; i--) {
    ans += res[i - 1] - res[i] - 1;
  }
  ans += res.back();
  cout << ans << endl;
}
int main() {
  srand((int)time(NULL));
  LoAd();
  SoLvE();
  return 0;
}
