#include <bits/stdc++.h>
using namespace std;
map<int, int> cnt;
int main() {
  int n;
  scanf("%d", &n);
  vector<int> lis;
  int data[n + 10];
  for (int i = 0; i < n; i++) {
    scanf("%d", &data[i]);
    cnt[data[i]]++;
    lis.push_back(data[i]);
  }
  sort(data, data + n);
  int pos = 0;
  int ans = 0;
  if (n == 1 && data[0] == 0) {
    cout << 0;
    return 0;
  }
  sort(lis.begin(), lis.end());
  reverse(lis.begin(), lis.end());
  set<int> Set;
  int ma = -1;
  while (!lis.empty()) {
    int cnt = 0;
    int val = lis.back();
    while (!lis.empty() && lis.back() == val) {
      cnt++;
      lis.pop_back();
    }
    if (cnt & 1) {
      Set.insert(val);
      ma = max(ma, val);
    }
    int x = cnt / 2;
    for (int i = 0; i < x; i++) lis.push_back(val + 1);
  }
  ma++;
  cout << ma - Set.size() << 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> cnt;
int main() {
  int n;
  scanf("%d", &n);
  vector<int> lis;
  int data[n + 10];
  for (int i = 0; i < n; i++) {
    scanf("%d", &data[i]);
    cnt[data[i]]++;
    lis.push_back(data[i]);
  }
  sort(data, data + n);
  int pos = 0;
  int ans = 0;
  if (n == 1 && data[0] == 0) {
    cout << 0;
    return 0;
  }
  sort(lis.begin(), lis.end());
  reverse(lis.begin(), lis.end());
  set<int> Set;
  int ma = -1;
  while (!lis.empty()) {
    int cnt = 0;
    int val = lis.back();
    while (!lis.empty() && lis.back() == val) {
      cnt++;
      lis.pop_back();
    }
    if (cnt & 1) {
      Set.insert(val);
      ma = max(ma, val);
    }
    int x = cnt / 2;
    for (int i = 0; i < x; i++) lis.push_back(val + 1);
  }
  ma++;
  cout << ma - Set.size() << endl;
  return 0;
}

```