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;
long long const mod = 1000000007;
long long const md = 998244353;
long long mypowr(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long mypow(long long a, long long b) {
  long long res = 1;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    a = a * a;
  }
  return res;
}
ifstream in("txt.in");
ofstream out("txt.out");
long long f[100005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(15);
  long long n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> f[i];
  }
  long long space = n;
  for (int i = 1; i <= n; ++i) {
    int j = i + 1;
    long long cnt = 0;
    while (f[j] == f[j - 1] && j <= n) {
      ++j;
    }
    --j;
    cnt = j - i + 1;
    if (cnt % 2) {
      ++i;
    }
    cnt /= 2;
    space -= cnt;
    i += cnt;
    for (int t = i; t <= i + cnt - 1; ++t) ++f[t];
    --i;
  }
  cout << f[n] - space + 1 << '\n';
  return 0;
}
