#include <bits/stdc++.h>
using namespace std;
int num[100005];
int main() {
  int n;
  while (cin >> n) {
    for (int i = 0; i < n; ++i) cin >> num[i];
    int ans = 0;
    int last = 0;
    int cur = num[0];
    int curcnt = 0;
    for (int i = 0; i < n; ++i) {
      if (num[i] == cur) {
        curcnt++;
        continue;
      }
      if (curcnt % 2 == 1) {
        ans += cur - last;
        cur = cur + 1;
        curcnt /= 2;
      } else {
        ans += cur - last + 1;
        cur = cur + 1;
        curcnt /= 2;
      }
      last = cur;
      if (curcnt != 0) {
        i--;
      } else {
        cur = num[i];
        curcnt = 1;
      }
    }
    while (curcnt != 0) {
      if (curcnt % 2 == 1) {
        ans += cur - last;
        cur = cur + 1;
        curcnt /= 2;
      } else {
        ans += cur - last + 1;
        cur = cur + 1;
        curcnt /= 2;
      }
      last = cur;
    }
    cout << ans << endl;
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 num[100005];
int main() {
  int n;
  while (cin >> n) {
    for (int i = 0; i < n; ++i) cin >> num[i];
    int ans = 0;
    int last = 0;
    int cur = num[0];
    int curcnt = 0;
    for (int i = 0; i < n; ++i) {
      if (num[i] == cur) {
        curcnt++;
        continue;
      }
      if (curcnt % 2 == 1) {
        ans += cur - last;
        cur = cur + 1;
        curcnt /= 2;
      } else {
        ans += cur - last + 1;
        cur = cur + 1;
        curcnt /= 2;
      }
      last = cur;
      if (curcnt != 0) {
        i--;
      } else {
        cur = num[i];
        curcnt = 1;
      }
    }
    while (curcnt != 0) {
      if (curcnt % 2 == 1) {
        ans += cur - last;
        cur = cur + 1;
        curcnt /= 2;
      } else {
        ans += cur - last + 1;
        cur = cur + 1;
        curcnt /= 2;
      }
      last = cur;
    }
    cout << ans << endl;
  }
}

```