#include <bits/stdc++.h>
const int NMAX = 100005;
int a[NMAX], n, tmax, tcnt;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[n - i - 1]);
  while (n > 0) {
    int j = n - 1, cnt;
    while (j > 0 && a[n - 1] == a[j - 1]) --j;
    cnt = n - j;
    if (cnt % 2 == 1) {
      tcnt++, n--;
      if (a[n] > tmax) tmax = a[n];
    }
    cnt /= 2;
    n -= cnt;
    for (int i = 0; i < cnt; i++) a[n - 1 - i]++;
  }
  printf("%d\n", tmax - tcnt + 1);
}

### 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>
const int NMAX = 100005;
int a[NMAX], n, tmax, tcnt;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[n - i - 1]);
  while (n > 0) {
    int j = n - 1, cnt;
    while (j > 0 && a[n - 1] == a[j - 1]) --j;
    cnt = n - j;
    if (cnt % 2 == 1) {
      tcnt++, n--;
      if (a[n] > tmax) tmax = a[n];
    }
    cnt /= 2;
    n -= cnt;
    for (int i = 0; i < cnt; i++) a[n - 1 - i]++;
  }
  printf("%d\n", tmax - tcnt + 1);
}

```