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 MAXN(100001);
const int MAXE(5000010);
const int MAXK(100010);
const int MAXC(2);
const int INF((INT_MAX - 1) / 2);
const int BASE(20);
const double LIM(1e50);
const double EPS(1e-5);
const double PER(0.02);
const double MUL(3.5);
int IT;
template <typename T>
inline bool checkmax(T &a, const T &b) {
  return b > a ? ((a = b), true) : false;
}
template <typename T>
inline bool checkmin(T &a, const T &b) {
  return b < a ? ((a = b), true) : false;
}
template <typename T>
inline T ABS(T a) {
  return a < 0 ? -a : a;
}
inline int dis(int x1, int y1, int x2, int y2) {
  return ABS(x1 - x2) + ABS(y1 - y2);
}
struct BIT {
  long long arr[200001], sz;
  int lb(int i) { return i & -i; }
  void ini(int sz) {
    this->sz = sz;
    memset(arr + 1, 0, sizeof(arr[0]) * sz);
  }
  void add(int i, long long v) {
    for (; i <= sz; i += lb(i)) arr[i] += v;
  }
  int query(int i) {
    int ret = 0;
    for (; i > 0; i -= lb(i)) ret += arr[i];
    return ret;
  }
};
int gcd(int a, int b) {
  while (b) {
    int t = a % b;
    a = b;
    b = t;
  }
  return a;
}
int arr[MAXN], qua[MAXN];
int main() {
  int n;
  long long rem = 0, cur = 0, ans = 0;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> arr[i];
  int tn = 0, pt = 1;
  while (pt <= n) {
    int temp = arr[pt++];
    arr[++tn] = temp;
    qua[tn] = 1;
    while (pt <= n && arr[pt] == temp) {
      ++qua[tn];
      ++pt;
    }
  }
  pt = 1;
  while (rem > 0 || pt <= tn) {
    if (pt <= n && cur == arr[pt]) rem += qua[pt++];
    if (rem == 0) {
      if (pt <= n) {
        ans += arr[pt] - cur;
        cur = arr[pt];
        rem = qua[pt++];
      }
    }
    ans += (rem & 1) ^ 1;
    rem /= 2;
    ++cur;
  }
  cout << ans << "\n";
  return 0;
}
