#include <bits/stdc++.h>
using namespace std;
map<long long, int> x;
int main() {
  long long n, a[100005];
  set<long long> z;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int i = 0;
  while (i < n) {
    int j = i;
    while (j < n && a[i] == a[j]) j++;
    x[a[i]] = j - i;
    z.insert(a[i]);
    i = j;
  }
  set<long long>::iterator it;
  int ans = 0;
  it = z.begin();
  long long num;
  for (; it != z.end(); it++) {
    num = *it;
    int ct = x[num];
    if (ct & 1) ans++, ct--;
    ;
    int k = 0;
    while (ct) {
      if (ct & 1) {
        if (x.count(num + k))
          x[num + k] += 1;
        else {
          x[num + k] = 1;
          z.insert(num + k);
        }
      }
      ct /= 2;
      k++;
    }
  }
  cout << num - ans + 1 << 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;
map<long long, int> x;
int main() {
  long long n, a[100005];
  set<long long> z;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int i = 0;
  while (i < n) {
    int j = i;
    while (j < n && a[i] == a[j]) j++;
    x[a[i]] = j - i;
    z.insert(a[i]);
    i = j;
  }
  set<long long>::iterator it;
  int ans = 0;
  it = z.begin();
  long long num;
  for (; it != z.end(); it++) {
    num = *it;
    int ct = x[num];
    if (ct & 1) ans++, ct--;
    ;
    int k = 0;
    while (ct) {
      if (ct & 1) {
        if (x.count(num + k))
          x[num + k] += 1;
        else {
          x[num + k] = 1;
          z.insert(num + k);
        }
      }
      ct /= 2;
      k++;
    }
  }
  cout << num - ans + 1 << endl;
}

```