#include <bits/stdc++.h>
using namespace std;
int n, M;
int flag[100100];
int x;
long long ans;
int t;
long long pos;
int main() {
  while (~scanf("%d", &n)) {
    memset(flag, 0, sizeof(flag));
    ans = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &flag[i]);
    }
    sort(flag + 1, flag + n + 1);
    t = 0;
    for (int i = 1; i <= n; i++) {
      if (t == 0) {
        pos = flag[i];
        t = 1;
      } else {
        if (flag[i] == pos)
          t++;
        else {
          if (t % 2) ans++;
          t >>= 1;
          pos++;
          i--;
        }
      }
    }
    while (t) {
      if (t & 1) ans++;
      t >>= 1;
      pos++;
    }
    ans = pos - ans;
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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>
using namespace std;
int n, M;
int flag[100100];
int x;
long long ans;
int t;
long long pos;
int main() {
  while (~scanf("%d", &n)) {
    memset(flag, 0, sizeof(flag));
    ans = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &flag[i]);
    }
    sort(flag + 1, flag + n + 1);
    t = 0;
    for (int i = 1; i <= n; i++) {
      if (t == 0) {
        pos = flag[i];
        t = 1;
      } else {
        if (flag[i] == pos)
          t++;
        else {
          if (t % 2) ans++;
          t >>= 1;
          pos++;
          i--;
        }
      }
    }
    while (t) {
      if (t & 1) ans++;
      t >>= 1;
      pos++;
    }
    ans = pos - ans;
    printf("%I64d\n", ans);
  }
  return 0;
}

```