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.
在这里充分用到了set去重的功能。要知道2^n +2^n= 2^(n+1).先输入x，如果发现x在集合里面，就要把x去掉，然后将x+1存到里面，如果x+1也在里面的话，就去掉将x+2存到里面，这是一个循环的过程。代码如下：
#include<iostream>
#include<cstring>
#include<set>
#include<cmath>
using namespace std;
set<int>num;
int n;
int main()
{
while(scanf("%d",&n)!=EOF)
{
int a;
num.clear();
int maxn=-0x3f3f3f3f;
while(n--)
{
scanf("%d",&a);
while(num.count(a))
{
num.erase(a);
a++;
}
num.insert(a);
maxn=max(maxn,a);
}
printf("%d\n",maxn+1-num.size());
}
}
努力加油a啊，(o)/~