#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  while (cin >> n) {
    vector<long long> a(n, 0);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    int cnt = 0;
    int carry = 0;
    int maxb = 0;
    int filled = 0;
    int i = 0;
    while (i < n) {
      int j = i;
      while (a[j] == a[i] && j < n) j++;
      cnt = j - i;
      filled += cnt % 2 == 0 ? 0 : 1;
      carry = cnt / 2;
      i = j - carry;
      for (int k = i; k < j; ++k) a[k]++;
    }
    maxb = a[n - 1];
    cout << maxb + 1 - filled << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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 main() {
  int n;
  while (cin >> n) {
    vector<long long> a(n, 0);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    int cnt = 0;
    int carry = 0;
    int maxb = 0;
    int filled = 0;
    int i = 0;
    while (i < n) {
      int j = i;
      while (a[j] == a[i] && j < n) j++;
      cnt = j - i;
      filled += cnt % 2 == 0 ? 0 : 1;
      carry = cnt / 2;
      i = j - carry;
      for (int k = i; k < j; ++k) a[k]++;
    }
    maxb = a[n - 1];
    cout << maxb + 1 - filled << endl;
  }
  return 0;
}

```