#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 10, MAX = 2e5 + 1e4, MOD = 1e9 + 7;
void OUT(long double o, int x) {
  cout << fixed << setprecision(x) << o;
  return;
}
long long a[MAX];
set<long long> st;
map<long long, long long> mp;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
    st.insert(a[i]);
  }
  long long m = 0, x = 0;
  for (long long i : st) {
    if (mp[i] > 1) st.insert(i + 1);
    mp[i + 1] += mp[i] / 2;
    mp[i] = mp[i] % 2;
    m = i;
    if (mp[i] != 0) x++;
  }
  cout << m + 1 - x;
  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;
const long long INF = 1e9 + 10, MAX = 2e5 + 1e4, MOD = 1e9 + 7;
void OUT(long double o, int x) {
  cout << fixed << setprecision(x) << o;
  return;
}
long long a[MAX];
set<long long> st;
map<long long, long long> mp;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
    st.insert(a[i]);
  }
  long long m = 0, x = 0;
  for (long long i : st) {
    if (mp[i] > 1) st.insert(i + 1);
    mp[i + 1] += mp[i] / 2;
    mp[i] = mp[i] % 2;
    m = i;
    if (mp[i] != 0) x++;
  }
  cout << m + 1 - x;
  return 0;
}

```