#include <bits/stdc++.h>
using namespace std;
const int MaxN = 1e5 + 5;
int box[MaxN], seq[MaxN];
int ans, n;
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &box[i]);
    int l = 1, r = 0;
    for (int i = n; i >= 1; i--) {
      while (r >= l && box[i] == seq[r]) {
        r--;
        box[i]++;
      }
      seq[++r] = box[i];
    }
    int temp = 0;
    ans = 0;
    for (int i = r; i >= 1; i--) {
      if (temp == seq[i]) {
        temp++;
      } else {
        ans += seq[i] - temp;
        temp = seq[i] + 1;
      }
    }
    printf("%d\n", ans);
    memset(seq, 0, sizeof(seq));
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
const int MaxN = 1e5 + 5;
int box[MaxN], seq[MaxN];
int ans, n;
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%d", &box[i]);
    int l = 1, r = 0;
    for (int i = n; i >= 1; i--) {
      while (r >= l && box[i] == seq[r]) {
        r--;
        box[i]++;
      }
      seq[++r] = box[i];
    }
    int temp = 0;
    ans = 0;
    for (int i = r; i >= 1; i--) {
      if (temp == seq[i]) {
        temp++;
      } else {
        ans += seq[i] - temp;
        temp = seq[i] + 1;
      }
    }
    printf("%d\n", ans);
    memset(seq, 0, sizeof(seq));
  }
  return 0;
}

```