#include <bits/stdc++.h>
using namespace std;
int a[100010];
map<int, int> m;
map<int, int>::iterator it, it2;
int main() {
  int n;
  scanf("%d", &n);
  int has = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (m.find(a[i]) != m.end()) {
      m[a[i]]++;
      m[a[i] + 1] += m[a[i]] / 2;
      m[a[i]] = m[a[i]] % 2;
      if (m[a[i]] == 0) m.erase(a[i]);
    } else
      m[a[i]] = 1;
  }
  int max = 0;
  for (it = m.begin(); it != m.end(); it++) {
    if (it->second == 1) {
      max = it->first;
      has++;
    } else {
      if (it->second % 2 == 1) has++;
      m[it->first + 1] += it->second / 2;
    }
  }
  cout << max + 1 - has << endl;
  return 0;
}

### Prompt

Please formulate 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 a[100010];
map<int, int> m;
map<int, int>::iterator it, it2;
int main() {
  int n;
  scanf("%d", &n);
  int has = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (m.find(a[i]) != m.end()) {
      m[a[i]]++;
      m[a[i] + 1] += m[a[i]] / 2;
      m[a[i]] = m[a[i]] % 2;
      if (m[a[i]] == 0) m.erase(a[i]);
    } else
      m[a[i]] = 1;
  }
  int max = 0;
  for (it = m.begin(); it != m.end(); it++) {
    if (it->second == 1) {
      max = it->first;
      has++;
    } else {
      if (it->second % 2 == 1) has++;
      m[it->first + 1] += it->second / 2;
    }
  }
  cout << max + 1 - has << endl;
  return 0;
}

```