#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e3 + 7;
const int mod = 1e9 + 7;
map<int, int> mp;
int n, x, ma, ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    mp[x]++;
  }
  for (auto p : mp) {
    int x = p.first, y = p.second;
    for (int i = 1; i <= 20; i++) {
      if ((y >> i) & 1) {
        mp[x + i]++;
        mp[x] -= (1 << i);
      }
    }
  }
  for (auto p : mp) {
    int x = p.first, y = p.second;
    for (int i = 1; i <= 20; i++) {
      if ((y >> i) & 1) {
        mp[x + i]++;
        mp[x] -= (1 << i);
      }
    }
  }
  for (auto p : mp) {
    int x = p.first, y = p.second;
    ;
    if (y != 0) {
      ans++;
      ma = max(ma, x);
    }
  }
  printf("%d\n", ma + 1 - ans);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int maxn = 2e3 + 7;
const int mod = 1e9 + 7;
map<int, int> mp;
int n, x, ma, ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    mp[x]++;
  }
  for (auto p : mp) {
    int x = p.first, y = p.second;
    for (int i = 1; i <= 20; i++) {
      if ((y >> i) & 1) {
        mp[x + i]++;
        mp[x] -= (1 << i);
      }
    }
  }
  for (auto p : mp) {
    int x = p.first, y = p.second;
    for (int i = 1; i <= 20; i++) {
      if ((y >> i) & 1) {
        mp[x + i]++;
        mp[x] -= (1 << i);
      }
    }
  }
  for (auto p : mp) {
    int x = p.first, y = p.second;
    ;
    if (y != 0) {
      ans++;
      ma = max(ma, x);
    }
  }
  printf("%d\n", ma + 1 - ans);
}

```