#include <bits/stdc++.h>
using namespace std;
void _fill_int(int* p, int val, int rep) {
  int i;
  for (i = 0; i < rep; i++) p[i] = val;
}
int GETi() {
  int i;
  scanf("%d", &i);
  return i;
}
template <class T>
T sqr(T val) {
  return val * val;
}
int N;
signed long long A[1000001];
void solve() {
  int f, r, i, j, k, l, x, y, ma, nt;
  signed long long res = 0, num = 0, digit = 0;
  N = GETi();
  for (i = 0; i < N; i++) A[i] = GETi();
  j = 0;
  while (j < N || num) {
    while (j < N && digit == A[j]) {
      num++;
      j++;
    }
    if (j < N && num == 0) {
      res += A[j] - digit;
      digit = A[j];
    } else {
      digit++;
      res += 1 - (num & 1);
      num >>= 1;
    }
  }
  cout << res << endl;
  return;
}
int main(int argc, char** argv) {
  if (argc > 1) freopen(argv[1], "r", stdin);
  solve();
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void _fill_int(int* p, int val, int rep) {
  int i;
  for (i = 0; i < rep; i++) p[i] = val;
}
int GETi() {
  int i;
  scanf("%d", &i);
  return i;
}
template <class T>
T sqr(T val) {
  return val * val;
}
int N;
signed long long A[1000001];
void solve() {
  int f, r, i, j, k, l, x, y, ma, nt;
  signed long long res = 0, num = 0, digit = 0;
  N = GETi();
  for (i = 0; i < N; i++) A[i] = GETi();
  j = 0;
  while (j < N || num) {
    while (j < N && digit == A[j]) {
      num++;
      j++;
    }
    if (j < N && num == 0) {
      res += A[j] - digit;
      digit = A[j];
    } else {
      digit++;
      res += 1 - (num & 1);
      num >>= 1;
    }
  }
  cout << res << endl;
  return;
}
int main(int argc, char** argv) {
  if (argc > 1) freopen(argv[1], "r", stdin);
  solve();
  return 0;
}

```