【CodeForces 570B】Simple Game（水）



Simple Game

One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from1 to n. Let's assume that Misha chose numberm, and Andrew chose number a.

Then, by using a random generator they choose a random integer c in the range between 1 and n (any integer from 1 to n is chosen with the same probability), after which the winner is the player, whose number was closer toc. The boys agreed that if m and a are located on the same distance fromc, Misha wins.

Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and numbern. You need to determine which value of a Andrew must choose, so that the probability of his victory is the highest possible.

More formally, you need to find such integer a (1 ≤ a ≤ n), that the probability that is maximal, wherec is the equiprobably chosen integer from 1 to n (inclusive).

Input

The first line contains two integers n andm (1 ≤ m ≤ n ≤ 10⁹) — the range of numbers in the game, and the number selected by Misha respectively.

Output

Print a single number — such value a, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them.

Example

Input

3 1

Output

2

Input

4 3

Output

2

Note

In the first sample test: Andrew wins if c is equal to2 or 3. The probability that Andrew wins is2 / 3. If Andrew chooses a = 3, the probability of winning will be1 / 3. If a = 1, the probability of winning is0.

In the second sample test: Andrew wins if c is equal to1 and 2. The probability that Andrew wins is1 / 2. For other choices of a the probability of winning is less.

题目大意：n个数，两个人分别选两个数，所有数中离谁的数近的多，谁赢，第一个人选了m，问第二个选几能赢

思路：概率问题，和中间的值做比较

#include<iostream>using namespace std;int main(){    int n,m;    cin>>n>>m;    int mid=(n+1)/2;    if(m==1 && n==1) cout<<"1"<<endl;    else if(m < mid) cout<<m+1<<endl;    else if(m > mid) cout<<m-1<<endl;    else {        if(n%2) cout<<m-1<<endl;        else cout<<m+1<<endl;    }    return 0;}