#include <bits/stdc++.h>
using namespace std;
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
int main() {
  int n;
  scanf("%d", &n);
  int a[n + 1];
  vector<long long int> v;
  map<int, int> ase;
  int mx = -1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    int now = a[i];
    ase[now]++;
    mx = max(now, mx);
    while (ase[now] == 2) {
      ase.erase(now);
      ase[++now]++;
      mx = max(now, mx);
    }
  }
  cout << (mx + 1 - ase.size());
  return 0;
}

### Prompt

Develop a solution in Cpp 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 Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
int main() {
  int n;
  scanf("%d", &n);
  int a[n + 1];
  vector<long long int> v;
  map<int, int> ase;
  int mx = -1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    int now = a[i];
    ase[now]++;
    mx = max(now, mx);
    while (ase[now] == 2) {
      ase.erase(now);
      ase[++now]++;
      mx = max(now, mx);
    }
  }
  cout << (mx + 1 - ase.size());
  return 0;
}

```