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;
const long long M = 1000000007;
const long long MM = 998244353;
long long begtime = clock();
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <typename T, typename F>
void chmax(T& a, F b) {
  if (b > a) a = b;
}
template <typename T, typename F>
void chmin(T& a, F b) {
  if (b < a) a = b;
}
long long powM(long long a, long long b, long long m) {
  if (a <= 0) return 0;
  a %= m;
  long long ans = 1LL;
  while (b) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return ans;
}
long long poww(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a;
    a = a * a;
    b >>= 1;
  }
  return ans;
}
string tostring(long long x) {
  stringstream sss;
  sss << x;
  string ans = sss.str();
  return ans;
}
const long long N = 1e5 + 5;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n;
  cin >> n;
  long long c = 0;
  vector<long long> a(n);
  vector<long long> w, b;
  w.push_back(0);
  b.push_back(0);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (i && a[i] != a[i - 1]) {
      w.push_back(c);
      b.push_back(a[i - 1]);
      c = 1;
    } else
      c++;
  }
  w.push_back(c);
  b.push_back(a[n - 1]);
  n = ((long long)w.size());
  long long ans = 0;
  for (long long i = 0; i < n - 1; i++) {
    long long x = 1e9;
    if (b[i + 1] - b[i] < 20) {
      x = 1ll << (b[i + 1] - b[i]);
    }
    long long y = w[i] % (x);
    ans += b[i + 1] - b[i] - __builtin_popcountll(y);
    w[i + 1] += w[i] / x;
  }
  while (w[n - 1]) {
    if (w[n - 1] % 2 == 0) ans++;
    w[n - 1] /= 2;
  }
  cout << ans;
  return 0;
}
