#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 5000 + 1, INF = 1e9, MOD = 1e9 + 7;
long long n, ans;
map<long long, long long> m;
set<long long> s;
void upd(long long t) {
  if (m[t] % 2) {
    if (m[t] <= 1) return;
    m[t + 1] += m[t] / 2;
    m[t] = 1;
    s.insert(t + 1);
  } else {
    s.erase(t);
    s.insert(t + 1);
    m[t + 1] += m[t] / 2;
    m[t] = 0;
  }
  upd(t + 1);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    long long a;
    cin >> a;
    if (s.size() && !s.count(a)) {
      long long t = *s.rbegin();
      upd(t);
    }
    s.insert(a);
    m[a]++;
    if (i == n - 1) {
      upd(a);
    }
  }
  while (s.size()) {
    long long x = *s.rbegin();
    s.erase(x);
    if (s.size() > 0) {
      long long t = *s.rbegin();
      ans += (x - t - 1);
    } else {
      ans += x;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 long long MAXN = 5000 + 1, INF = 1e9, MOD = 1e9 + 7;
long long n, ans;
map<long long, long long> m;
set<long long> s;
void upd(long long t) {
  if (m[t] % 2) {
    if (m[t] <= 1) return;
    m[t + 1] += m[t] / 2;
    m[t] = 1;
    s.insert(t + 1);
  } else {
    s.erase(t);
    s.insert(t + 1);
    m[t + 1] += m[t] / 2;
    m[t] = 0;
  }
  upd(t + 1);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    long long a;
    cin >> a;
    if (s.size() && !s.count(a)) {
      long long t = *s.rbegin();
      upd(t);
    }
    s.insert(a);
    m[a]++;
    if (i == n - 1) {
      upd(a);
    }
  }
  while (s.size()) {
    long long x = *s.rbegin();
    s.erase(x);
    if (s.size() > 0) {
      long long t = *s.rbegin();
      ans += (x - t - 1);
    } else {
      ans += x;
    }
  }
  cout << ans << endl;
  return 0;
}

```