#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int n;
set<int> h, ans;
map<int, int> res;
void run() {
  int x, cnt;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    h.insert(x);
    res[x]++;
  }
  int mm = -1, size = 0;
  for (set<int>::iterator it = h.begin(); it != h.end(); it++) {
    x = *it, cnt = res[x];
    mm = x;
    size++;
    if (cnt >= 2) {
      if (cnt % 2 == 0) size--;
      h.insert(x + 1);
      res[x + 1] += cnt / 2;
    }
  }
  printf("%d\n", mm + 1 - size);
}
int main() {
  run();
  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;
const int N = 100005;
int n;
set<int> h, ans;
map<int, int> res;
void run() {
  int x, cnt;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    h.insert(x);
    res[x]++;
  }
  int mm = -1, size = 0;
  for (set<int>::iterator it = h.begin(); it != h.end(); it++) {
    x = *it, cnt = res[x];
    mm = x;
    size++;
    if (cnt >= 2) {
      if (cnt % 2 == 0) size--;
      h.insert(x + 1);
      res[x + 1] += cnt / 2;
    }
  }
  printf("%d\n", mm + 1 - size);
}
int main() {
  run();
  return 0;
}

```