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.
#include <bits/stdc++.h>
using namespace std;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
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) {
  long long g = gcd(a, b);
  long long ans = (a * b) / g;
  return ans;
}
long long add(long long a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  return a;
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) {
    a += 1000000007;
  }
  return a;
}
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
vector<long long> primes;
bool prime[10005];
void seive() {
  memset(prime, 1, sizeof(prime));
  prime[0] = 0;
  prime[1] = 0;
  for (long long i = 2; i <= 10000; i++) {
    if (prime[i] == 1) {
      for (long long j = i * i; j <= 10000; j += i) prime[j] = 0;
    }
  }
}
long long power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans % 1000000007 * a % 1000000007) % 1000000007;
    a = (a * a) % 1000000007;
    b = b / 2;
  }
  return ans;
}
template <typename T>
std::string NumberToString(T Number) {
  std::ostringstream ss;
  ss << Number;
  return ss.str();
}
string num_to_bits(long long n) {
  string ans = "";
  while (n) {
    ans += (n % 2) + '0';
    n /= 2;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
long long bits_to_num(string s) {
  long long ans = 0;
  for (long long i = 0; i < s.size(); i++) {
    long long x = s.size() - i - 1;
    long long val = pow(2LL, x);
    if (s[i] == '1') {
      ans = ans + val;
    }
  }
  return ans;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y);
long long modInverse(long long b, long long m) {
  long long x, y;
  long long g = gcdExtended(b, m, &x, &y);
  if (g != 1) return -1;
  return (x % m + m) % m;
}
long long modDivide(long long a, long long b, long long m) {
  a = a % m;
  long long inv = modInverse(b, m);
  return (inv * a) % m;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n], i, j;
  map<long long, long long> mp;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  vector<long long> v;
  for (auto it : mp) {
    long long val = it.first, count = it.second;
    if (count % 2 == 1) {
      v.push_back(val);
    }
    if (count / 2 != 0) {
      mp[val + 1] += count / 2;
    }
  }
  cout << v[v.size() - 1] - v.size() + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  bool codechef = 0;
  long long t = 1;
  if (codechef) {
    cin >> t;
  }
  while (t--) {
    solve();
  }
  return 0;
}
