#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 2;
const int mod = 1e9 + 7;
const double eps = 1e-8;
inline int read() {
  int x;
  scanf("%d", &x);
  return x;
}
long long a[N], b[N];
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
  }
  while (1) {
    bool ok = 0;
    int id = 0;
    for (int i = 1; i <= n; i++) {
      if (i < n && a[i] == a[i + 1]) {
        b[++id] = a[i] + 1;
        ok = 1;
        i++;
      } else
        b[++id] = a[i];
    }
    n = id;
    for (int i = 1; i <= id; i++) a[i] = b[i];
    if (!ok) break;
    sort(a + 1, a + n + 1);
  }
  long long ans = 0;
  a[0] = -1;
  for (int i = 1; i <= n; i++) {
    ans += a[i] - a[i - 1] - 1;
  }
  cout << ans;
}

### 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>
using namespace std;
const int N = 1e6 + 2;
const int mod = 1e9 + 7;
const double eps = 1e-8;
inline int read() {
  int x;
  scanf("%d", &x);
  return x;
}
long long a[N], b[N];
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
  }
  while (1) {
    bool ok = 0;
    int id = 0;
    for (int i = 1; i <= n; i++) {
      if (i < n && a[i] == a[i + 1]) {
        b[++id] = a[i] + 1;
        ok = 1;
        i++;
      } else
        b[++id] = a[i];
    }
    n = id;
    for (int i = 1; i <= id; i++) a[i] = b[i];
    if (!ok) break;
    sort(a + 1, a + n + 1);
  }
  long long ans = 0;
  a[0] = -1;
  for (int i = 1; i <= n; i++) {
    ans += a[i] - a[i - 1] - 1;
  }
  cout << ans;
}

```