#include <bits/stdc++.h>
using namespace std;
long long factorial(long long n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool sortByVal(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  if (a.first == b.first)
    return a.second < b.second;
  else
    return a.first > b.first;
}
long long Powb(long long b, long long n) {
  if (n == 0) return 1LL;
  if (n == 1) return b;
  long long temp = Powb(b, n / 2);
  if (n % 2 == 0) {
    return (temp * temp) % 1000000007LL;
  } else {
    return (b * ((temp * temp) % 1000000007LL)) % 1000000007LL;
  }
}
long long root(vector<long long> &a, long long i) {
  while (a[i] != i) {
    a[i] = a[a[i]];
    i = a[i];
  }
  return i;
}
void wunion(vector<long long> &a, vector<long long> &size, long long i,
            long long j) {
  long long root_i = root(a, i);
  long long root_j = root(a, j);
  if (root_i == root_j) return;
  if (size[root_i] < size[root_j]) {
    a[root_i] = a[root_j];
    size[root_j] += size[root_i];
    size[root_i] = 0;
  } else {
    a[root_j] = a[root_i];
    size[root_i] += size[root_j];
    size[root_j] = 0;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    long long i, j, m, n, k, sum = 0, c = 0, d = 0, e = 0, f = 0, g = 0, h = 0,
                             maxi = 0;
    cin >> n;
    multiset<long long> s;
    for (i = 0; i < n; i++) {
      cin >> c;
      s.insert(c);
    }
    auto now = s.begin();
    while (now != s.end() && s.size() > 1) {
      auto nxt = next(now);
      if (*nxt == *now) {
        c = *now;
        s.erase(now);
        now = s.erase(nxt);
        s.insert(c + 1);
      } else
        now++;
    }
    d = *prev(s.end());
    cout << d - s.size() + 1 << "\n";
  }
}

### Prompt

Please formulate 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;
long long factorial(long long n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool sortByVal(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  if (a.first == b.first)
    return a.second < b.second;
  else
    return a.first > b.first;
}
long long Powb(long long b, long long n) {
  if (n == 0) return 1LL;
  if (n == 1) return b;
  long long temp = Powb(b, n / 2);
  if (n % 2 == 0) {
    return (temp * temp) % 1000000007LL;
  } else {
    return (b * ((temp * temp) % 1000000007LL)) % 1000000007LL;
  }
}
long long root(vector<long long> &a, long long i) {
  while (a[i] != i) {
    a[i] = a[a[i]];
    i = a[i];
  }
  return i;
}
void wunion(vector<long long> &a, vector<long long> &size, long long i,
            long long j) {
  long long root_i = root(a, i);
  long long root_j = root(a, j);
  if (root_i == root_j) return;
  if (size[root_i] < size[root_j]) {
    a[root_i] = a[root_j];
    size[root_j] += size[root_i];
    size[root_i] = 0;
  } else {
    a[root_j] = a[root_i];
    size[root_i] += size[root_j];
    size[root_j] = 0;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    long long i, j, m, n, k, sum = 0, c = 0, d = 0, e = 0, f = 0, g = 0, h = 0,
                             maxi = 0;
    cin >> n;
    multiset<long long> s;
    for (i = 0; i < n; i++) {
      cin >> c;
      s.insert(c);
    }
    auto now = s.begin();
    while (now != s.end() && s.size() > 1) {
      auto nxt = next(now);
      if (*nxt == *now) {
        c = *now;
        s.erase(now);
        now = s.erase(nxt);
        s.insert(c + 1);
      } else
        now++;
    }
    d = *prev(s.end());
    cout << d - s.size() + 1 << "\n";
  }
}

```