#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const double infd = 2e+9;
const int infi = 1ll << 30;
const long long infl = 1ll << 60;
const int mod = 1e+9 + 7;
template <class T>
inline T sqr(T x) {
  return x * x;
}
map<int, bool> m;
void add(int x) {
  if (m[x]) {
    m[x] = false;
    add(x + 1);
  } else
    m[x] = true;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    add(x);
  }
  int cnt = 0, max_bit = 0;
  for (map<int, bool>::iterator it = m.begin(); it != m.end(); it++) {
    if (it->second) cnt++;
    max_bit = max(max_bit, it->first);
  }
  cout << max_bit - cnt + 1;
  return 0;
}

### 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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const double infd = 2e+9;
const int infi = 1ll << 30;
const long long infl = 1ll << 60;
const int mod = 1e+9 + 7;
template <class T>
inline T sqr(T x) {
  return x * x;
}
map<int, bool> m;
void add(int x) {
  if (m[x]) {
    m[x] = false;
    add(x + 1);
  } else
    m[x] = true;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    add(x);
  }
  int cnt = 0, max_bit = 0;
  for (map<int, bool>::iterator it = m.begin(); it != m.end(); it++) {
    if (it->second) cnt++;
    max_bit = max(max_bit, it->first);
  }
  cout << max_bit - cnt + 1;
  return 0;
}

```