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>
const int Len = 100000;
int n;
int A[Len + 5];
int B[Len + 5];
int C[Len + 5];
int nn;
int cc;
void solve() {
  cc = 0;
  nn = 0;
  B[0] = A[0];
  C[0] = 1;
  for (int i = 1; i < n; ++i)
    if (A[i] != A[i - 1]) {
      B[++nn] = A[i];
      C[nn] = 1;
    } else
      ++C[nn];
  int num0 = A[0], num1;
  if (C[0] & 1) {
    ++cc;
    num1 = C[0] - 1;
  } else
    num1 = C[0];
  for (int i = 1; i <= nn; ++i) {
    int sub = B[i] - B[i - 1];
    while (num1 && sub--) {
      if (num1 & 1) ++cc;
      num1 /= 2;
    }
    num0 = B[i];
    num1 = C[i] + num1;
    if (num1 & 1) {
      --num1;
      ++cc;
    }
  }
  while (num1) {
    if (num1 > 1) ++num0;
    if (num1 & 1) ++cc;
    num1 /= 2;
  }
  printf("%d\n", num0 + 1 - cc);
}
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; ++i) scanf("%d", &A[i]);
    solve();
  }
  return 0;
}
