#include <bits/stdc++.h>
using namespace std;
int a, an, sm;
long long b[100009], z;
long long p[6000009];
int main() {
  scanf("%d", &a);
  for (int i = 0; i < a; i++) scanf("%I64d", &b[i]);
  b[a] = (long long)1e10;
  int z = 1;
  p[0] = -1;
  an = 1;
  for (int i = 1; i <= a; i++) {
    if (b[i] != b[i - 1]) {
      while (b[i] != b[i - 1] && z > 0) {
        if (z % 2 == 1) p[an] = b[i - 1], an++;
        z /= 2;
        b[i - 1]++;
      }
      z = z + 1;
    } else
      z++;
  }
  for (int i = 1; i < an; i++) sm = sm + p[i] - p[i - 1] - 1;
  printf("%d", sm);
  getchar();
  getchar();
}

### Prompt

Generate 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;
int a, an, sm;
long long b[100009], z;
long long p[6000009];
int main() {
  scanf("%d", &a);
  for (int i = 0; i < a; i++) scanf("%I64d", &b[i]);
  b[a] = (long long)1e10;
  int z = 1;
  p[0] = -1;
  an = 1;
  for (int i = 1; i <= a; i++) {
    if (b[i] != b[i - 1]) {
      while (b[i] != b[i - 1] && z > 0) {
        if (z % 2 == 1) p[an] = b[i - 1], an++;
        z /= 2;
        b[i - 1]++;
      }
      z = z + 1;
    } else
      z++;
  }
  for (int i = 1; i < an; i++) sm = sm + p[i] - p[i - 1] - 1;
  printf("%d", sm);
  getchar();
  getchar();
}

```