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_MAX = 0x7FFFFFFF;
const int INF_MIN = -(1 << 30);
const double eps = 1e-10;
const double pi = acos(-1.0);
int toInt(string s) {
  istringstream sin(s);
  int t;
  sin >> t;
  return t;
}
template <class T>
string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
template <class T>
inline void CLR(priority_queue<T, vector<T>, greater<T> > &Q) {
  while (!Q.empty()) Q.pop();
}
inline int random(int l, int r) { return rand() % (r - l + 1) + l; }
int dir_4[4][2] = {{0, 1}, {-1, 0}, {0, -1}, {1, 0}};
int dir_8[8][2] = {{0, 1},  {-1, 1}, {-1, 0}, {-1, -1},
                   {0, -1}, {1, -1}, {1, 0},  {1, 1}};
deque<int> Q;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int tmp;
    scanf("%d", &tmp);
    Q.push_back(tmp);
  }
  int maxb = 0, cntb = 0;
  while (!Q.empty()) {
    int cnt = 0;
    int f = Q.front();
    while (!Q.empty() && f == Q.front()) {
      cnt++;
      Q.pop_front();
    }
    if (cnt % 2 == 1) {
      cntb++;
    }
    maxb = ((maxb) > (f) ? (maxb) : (f));
    while (cnt >= 2) {
      Q.push_front(f + 1);
      cnt -= 2;
    }
  }
  printf("%d", maxb + 1 - cntb);
}
