#include <bits/stdc++.h>
int a[100010];
void solve(int n) {
  int i, j, cnt, num, len, ans;
  ans = a[0];
  cnt = 0;
  for (i = 0; i < n; i = j) {
    for (j = i + 1; j < n && a[j] == a[i]; j++)
      ;
    len = j - i + cnt;
    cnt = 0;
    num = a[i];
    while (len > 0 && j < n && num < a[j]) {
      if ((len & 1) == 0) {
        ans++;
      }
      num++;
      len >>= 1;
    }
    if (j < n) {
      if (len == 0) {
        ans += (a[j] - num);
      } else {
        cnt = len;
      }
    } else {
      while (len > 0) {
        if ((len & 1) == 0) {
          ans++;
        }
        len >>= 1;
      }
    }
  }
  printf("%d\n", ans);
}
int main() {
  int n, i;
  while (scanf("%d", &n) == 1) {
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    solve(n);
  }
  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>
int a[100010];
void solve(int n) {
  int i, j, cnt, num, len, ans;
  ans = a[0];
  cnt = 0;
  for (i = 0; i < n; i = j) {
    for (j = i + 1; j < n && a[j] == a[i]; j++)
      ;
    len = j - i + cnt;
    cnt = 0;
    num = a[i];
    while (len > 0 && j < n && num < a[j]) {
      if ((len & 1) == 0) {
        ans++;
      }
      num++;
      len >>= 1;
    }
    if (j < n) {
      if (len == 0) {
        ans += (a[j] - num);
      } else {
        cnt = len;
      }
    } else {
      while (len > 0) {
        if ((len & 1) == 0) {
          ans++;
        }
        len >>= 1;
      }
    }
  }
  printf("%d\n", ans);
}
int main() {
  int n, i;
  while (scanf("%d", &n) == 1) {
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    solve(n);
  }
  return 0;
}

```