#include <bits/stdc++.h>
using namespace std;
int a[100005];
int main() {
  int n;
  scanf("%d", &n);
  int i;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  a[n] = -1;
  for (i = 0; i < n; i++) {
    int sum = 0;
    int j;
    for (j = i; a[j] == a[i]; j++) {
      sum++;
    }
    int k = j - sum / 2;
    for (; k < j; k++) {
      a[k] = a[i] + 1;
    }
    int temp = a[i];
    for (k = i; k < j - sum / 2; k++) {
      a[k] = -1;
    }
    if (sum % 2 == 1) a[j - sum / 2 - 1] = temp;
    i = j - sum / 2 - 1;
  }
  sort(a, a + n);
  int tot = a[n - 1];
  for (i = n - 2; i >= 0; i--) {
    if (a[i] >= 0) tot--;
  }
  printf("%d\n", tot);
  return 0;
}

### Prompt

In CPP, your task is to solve 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[100005];
int main() {
  int n;
  scanf("%d", &n);
  int i;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  a[n] = -1;
  for (i = 0; i < n; i++) {
    int sum = 0;
    int j;
    for (j = i; a[j] == a[i]; j++) {
      sum++;
    }
    int k = j - sum / 2;
    for (; k < j; k++) {
      a[k] = a[i] + 1;
    }
    int temp = a[i];
    for (k = i; k < j - sum / 2; k++) {
      a[k] = -1;
    }
    if (sum % 2 == 1) a[j - sum / 2 - 1] = temp;
    i = j - sum / 2 - 1;
  }
  sort(a, a + n);
  int tot = a[n - 1];
  for (i = n - 2; i >= 0; i--) {
    if (a[i] >= 0) tot--;
  }
  printf("%d\n", tot);
  return 0;
}

```