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 int inf = 1000000007;
const long long linf = 1000000000000000000LL;
const double eps = 0.000001;
const double pi = 3.14159265358979323846;
template <class T>
T abs(T k) {
  return k >= 0 ? k : -k;
}
template <class T>
T sqr(T n) {
  return n * n;
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T mod(T a, T b) {
  return (a % b + b) % b;
}
template <class T>
T powmod(T a, T b, T c) {
  return b ? sqr((long long)powmod(a, b >> 1, c)) % c * (b & 1 ? a : 1) % c : 1;
}
template <class T>
void mode(T &a, T b) {
  a = mod(a, b);
}
template <class T>
void maxe(T &a, T b) {
  a = max(a, b);
}
template <class T>
void mine(T &a, T b) {
  a = min(a, b);
}
void fileio_in_out() {
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
}
void fileio_txt() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
const int N = 1111111;
int test;
int n, m, k, ans;
int a[N];
int log2(int k) {
  int ret = -1;
  while (k) {
    k /= 2;
    ret++;
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  m = a[1];
  k = 1;
  ans = m;
  for (int i = 2; i <= n; i++) {
    if (m + log2(k) < a[i]) {
      while (k > 1) {
        m++;
        ans += (k & 1) ^ 1;
        k >>= 1;
      }
      ans += a[i] - m - 1;
      m = a[i];
      k = 1;
    } else {
      while (m < a[i]) {
        m++;
        ans += (k & 1) ^ 1;
        k >>= 1;
      }
      k++;
    }
  }
  while (k > 1) {
    ans += (k & 1) ^ 1;
    k >>= 1;
  }
  printf("%d\n", ans);
  return 0;
}
