#include <bits/stdc++.h>
long long mod = 1000000007;
using namespace std;
int n;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
map<int, int> M;
int main() {
  int i, j, fl, test;
  scanf("%d", &n);
  int left = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", &j);
    M[j]++;
  }
  int ans = 0;
  int curr = -1;
  for (auto it = M.begin(); it != M.end(); it++) {
    int tmp = it->first;
    int freq = it->second;
    if (freq % 2 == 0) {
      ans++;
    }
    ans = ans + (tmp - curr - 1);
    curr = tmp;
    if (freq != 1)
      M[tmp + 1] =
          M[tmp + 1] + ((freq - 1) % 2 == 0 ? (freq - 1) / 2 : (freq) / 2);
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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>
long long mod = 1000000007;
using namespace std;
int n;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
map<int, int> M;
int main() {
  int i, j, fl, test;
  scanf("%d", &n);
  int left = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", &j);
    M[j]++;
  }
  int ans = 0;
  int curr = -1;
  for (auto it = M.begin(); it != M.end(); it++) {
    int tmp = it->first;
    int freq = it->second;
    if (freq % 2 == 0) {
      ans++;
    }
    ans = ans + (tmp - curr - 1);
    curr = tmp;
    if (freq != 1)
      M[tmp + 1] =
          M[tmp + 1] + ((freq - 1) % 2 == 0 ? (freq - 1) / 2 : (freq) / 2);
  }
  printf("%d\n", ans);
  return 0;
}

```