#include <bits/stdc++.h>
using namespace std;
int n;
int lst = -1;
int res;
map<int, int> cnt[2];
bool ok(int i) {
  bool ret = true;
  for (map<int, int>::iterator it = cnt[i].begin(); it != cnt[i].end(); ++it) {
    int x = it->second;
    for (int j = 0; j < 32; ++j)
      if (x >> j & 1) ret &= ++cnt[i ^ 1][it->first + j] < 2;
  }
  cnt[i].clear();
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0, a; i < n; ++i) {
    cin >> a;
    ++cnt[0][a];
  }
  for (int i = 0; !ok(i & 1); ++i)
    ;
  int x = cnt[0].size() ? 0 : 1;
  int res = 0;
  int lst = -1;
  for (map<int, int>::iterator it = cnt[x].begin(); it != cnt[x].end(); ++it) {
    res += it->first - lst - 1;
    lst = it->first;
  }
  cout << res << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
int lst = -1;
int res;
map<int, int> cnt[2];
bool ok(int i) {
  bool ret = true;
  for (map<int, int>::iterator it = cnt[i].begin(); it != cnt[i].end(); ++it) {
    int x = it->second;
    for (int j = 0; j < 32; ++j)
      if (x >> j & 1) ret &= ++cnt[i ^ 1][it->first + j] < 2;
  }
  cnt[i].clear();
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0, a; i < n; ++i) {
    cin >> a;
    ++cnt[0][a];
  }
  for (int i = 0; !ok(i & 1); ++i)
    ;
  int x = cnt[0].size() ? 0 : 1;
  int res = 0;
  int lst = -1;
  for (map<int, int>::iterator it = cnt[x].begin(); it != cnt[x].end(); ++it) {
    res += it->first - lst - 1;
    lst = it->first;
  }
  cout << res << endl;
  return 0;
}

```