#include <bits/stdc++.h>
using namespace std;
int a[100010];
map<int, int> mp;
int main() {
  int n, i, j, k, m, x, y;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]]++;
  }
  for (i = 0; i < n; i++) {
    x = a[i];
    y = mp[x];
    while (y > 1) {
      mp[x] = y % 2;
      mp[x + 1] += y / 2;
      x++;
      y = mp[x];
    }
  }
  int last = 0;
  int cnt = 0;
  for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++) {
    x = it->first;
    y = it->second;
    if (y) {
      last = x;
      cnt++;
    }
  }
  printf("%d\n", last - cnt + 1);
  return 0;
}

### Prompt

Create a solution in cpp for 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;
int a[100010];
map<int, int> mp;
int main() {
  int n, i, j, k, m, x, y;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]]++;
  }
  for (i = 0; i < n; i++) {
    x = a[i];
    y = mp[x];
    while (y > 1) {
      mp[x] = y % 2;
      mp[x + 1] += y / 2;
      x++;
      y = mp[x];
    }
  }
  int last = 0;
  int cnt = 0;
  for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++) {
    x = it->first;
    y = it->second;
    if (y) {
      last = x;
      cnt++;
    }
  }
  printf("%d\n", last - cnt + 1);
  return 0;
}

```