#include <bits/stdc++.h>
using namespace std;
const int maxn = 100100;
long long a[maxn], b[maxn];
int n, top;
long long x;
int main() {
  cin >> n;
  top = 0;
  cin >> a[top];
  b[top] = 1;
  for (int i = 1; i < n; i++) {
    cin >> x;
    if (x != a[top]) {
      a[++top] = x;
      b[top] = 1;
    } else {
      b[top]++;
    }
  }
  long long num = 0;
  for (int i = 0; i < top; i++) {
    long long x = b[i];
    long long y = a[i + 1] - a[i];
    while (x) {
      if (x & 1) num++;
      x >>= 1;
      y--;
      if (!y) {
        b[i + 1] += x;
        break;
      }
    }
  }
  long long ans = a[top];
  if (b[top] & 1) num++;
  b[top] >>= 1;
  while (b[top]) {
    ans++;
    if (b[top] & 1) num++;
    b[top] >>= 1;
  }
  cout << ans - num + 1 << endl;
}

### Prompt

Your challenge is to write 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 maxn = 100100;
long long a[maxn], b[maxn];
int n, top;
long long x;
int main() {
  cin >> n;
  top = 0;
  cin >> a[top];
  b[top] = 1;
  for (int i = 1; i < n; i++) {
    cin >> x;
    if (x != a[top]) {
      a[++top] = x;
      b[top] = 1;
    } else {
      b[top]++;
    }
  }
  long long num = 0;
  for (int i = 0; i < top; i++) {
    long long x = b[i];
    long long y = a[i + 1] - a[i];
    while (x) {
      if (x & 1) num++;
      x >>= 1;
      y--;
      if (!y) {
        b[i + 1] += x;
        break;
      }
    }
  }
  long long ans = a[top];
  if (b[top] & 1) num++;
  b[top] >>= 1;
  while (b[top]) {
    ans++;
    if (b[top] & 1) num++;
    b[top] >>= 1;
  }
  cout << ans - num + 1 << endl;
}

```