#include <bits/stdc++.h>
using namespace std;
map<int, int> mp;
int main() {
  int i, j, k, l, n, m;
  vector<int> arr;
  int zerobit = 0;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> j;
    mp[j]++;
    if (mp[j] == 1) arr.push_back(j);
  }
  sort(arr.begin(), arr.end());
  int ex = 0, minus = 0;
  ;
  for (i = 0; i < arr.size(); i++) {
    while (ex && minus < arr[i]) {
      zerobit += !(ex & 1);
      ex >>= 1;
      minus++;
    }
    if (minus == arr[i]) {
      ex += mp[arr[i]];
    } else {
      zerobit += arr[i] - minus;
      minus = arr[i];
      ex = mp[arr[i]];
    }
  }
  while (ex) {
    zerobit += !(ex & 1);
    ex >>= 1;
  }
  cout << zerobit << endl;
  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>
using namespace std;
map<int, int> mp;
int main() {
  int i, j, k, l, n, m;
  vector<int> arr;
  int zerobit = 0;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> j;
    mp[j]++;
    if (mp[j] == 1) arr.push_back(j);
  }
  sort(arr.begin(), arr.end());
  int ex = 0, minus = 0;
  ;
  for (i = 0; i < arr.size(); i++) {
    while (ex && minus < arr[i]) {
      zerobit += !(ex & 1);
      ex >>= 1;
      minus++;
    }
    if (minus == arr[i]) {
      ex += mp[arr[i]];
    } else {
      zerobit += arr[i] - minus;
      minus = arr[i];
      ex = mp[arr[i]];
    }
  }
  while (ex) {
    zerobit += !(ex & 1);
    ex >>= 1;
  }
  cout << zerobit << endl;
  return 0;
}

```