开汽车找房子

The main street of Berland is a straight line with n houses built along it (n is an even number). The houses are located at both sides of the street. The houses with odd numbers are at one side of the street and are numbered from 1 ton - 1 in the order from the beginning of the street to the end (in the picture: from left to right). The houses with even numbers are at the other side of the street and are numbered from2 to n in the order from the end of the street to its beginning (in the picture: from right to left). The corresponding houses with even and odd numbers are strictly opposite each other, that is, house 1 is opposite house n, house 3 is opposite house n - 2, house 5 is opposite house n - 4 and so on.

Vasya needs to get to house number a as quickly as possible. He starts driving from the beginning of the street and drives his car to housea. To get from the beginning of the street to houses number1 and n, he spends exactly1 second. He also spends exactly one second to drive the distance between two neighbouring houses. Vasya can park at any side of the road, so the distance between the beginning of the street at the houses that stand opposite one another should be considered the same.

Your task is: find the minimum time Vasya needs to reach house a.

Input

The first line of the input contains two integers, n anda (1 ≤ a ≤ n ≤ 100 000) — the number of houses on the street and the number of the house that Vasya needs to reach, correspondingly. It is guaranteed that numbern is even.

Output

Print a single integer — the minimum time Vasya needs to get from the beginning of the street to housea.

Sample Input

Input

4 2

Output

2

Input

8 5

Output

3题目大意：偶数的房屋在街的另一边,编号从2到n的顺序从街上的开始(图中:从右到左)。奇数和偶数相应的房屋与严格彼此相反,也就是说,1是对面的房子n,n - 2 3对上门的房子5是对面的房子n - 4等等。Vasya需要尽快门牌号。他开始开车从一开始的街道和开车的房子。从一开始的街道房屋数量1,n,他花1秒。他还花了一个第二驱动两个邻近的房屋之间的距离。题解：
#include<stdio.h>int main(){int n,m,k;while(~scanf("%d%d",&n,&m)){if(m%2==0)printf("%d\n",n/2-m/2+1);//当要到达房屋为偶数时elseprintf("%d\n",m/2+1);//当要达到房屋为奇数时}}