#include <bits/stdc++.h>
using namespace std;
vector<int> a, b;
int main() {
  int n;
  scanf("%d", &n);
  a.resize(n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  while (1) {
    bool f = false;
    int i = 0;
    b.clear();
    int n = a.size();
    while (i < a.size()) {
      int c = i;
      while ((c < n) && (a[c] == a[i])) c++;
      int h = c - i;
      if (h & 1) b.push_back(a[i]);
      if (h >= 2) f = true;
      for (int j = (0); j < (h / 2); ++j) b.push_back(a[i] + 1);
      i = c;
    }
    if (!f) break;
    a = b;
  }
  int up = a[a.size() - 1];
  int res = up + 1 - a.size();
  cout << res << endl;
  return 0;
}

### 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;
vector<int> a, b;
int main() {
  int n;
  scanf("%d", &n);
  a.resize(n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  while (1) {
    bool f = false;
    int i = 0;
    b.clear();
    int n = a.size();
    while (i < a.size()) {
      int c = i;
      while ((c < n) && (a[c] == a[i])) c++;
      int h = c - i;
      if (h & 1) b.push_back(a[i]);
      if (h >= 2) f = true;
      for (int j = (0); j < (h / 2); ++j) b.push_back(a[i] + 1);
      i = c;
    }
    if (!f) break;
    a = b;
  }
  int up = a[a.size() - 1];
  int res = up + 1 - a.size();
  cout << res << endl;
  return 0;
}

```