#include <bits/stdc++.h>
using namespace std;
void func(void) {
  freopen("input.c", "r", stdin);
  freopen("output.c", "w", stdout);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, q, i, j = 0, temp, t, k, ans = 0, sum = 0, cnt = 0, m, fg = 0, x,
                     y, mx = 0, mx1 = 0, mn = 10000000000000000,
                     mn1 = 10000000000000000;
  map<long long, long long> a;
  cin >> n;
  while (n--) {
    cin >> x;
    a[x]++;
  }
  auto it = a.begin();
  for (; it != a.end(); it++) {
    mx = max(mx, it->first);
    cnt += it->second % 2;
    k = it->second / 2;
    if (k) {
      a[it->first + 1] += k;
    }
  }
  cout << mx + 1 - cnt << endl;
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void func(void) {
  freopen("input.c", "r", stdin);
  freopen("output.c", "w", stdout);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, q, i, j = 0, temp, t, k, ans = 0, sum = 0, cnt = 0, m, fg = 0, x,
                     y, mx = 0, mx1 = 0, mn = 10000000000000000,
                     mn1 = 10000000000000000;
  map<long long, long long> a;
  cin >> n;
  while (n--) {
    cin >> x;
    a[x]++;
  }
  auto it = a.begin();
  for (; it != a.end(); it++) {
    mx = max(mx, it->first);
    cnt += it->second % 2;
    k = it->second / 2;
    if (k) {
      a[it->first + 1] += k;
    }
  }
  cout << mx + 1 - cnt << endl;
  return 0;
}

```