Codeforces 570B Simple Game【思维】

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本文介绍了一个简单的游戏场景，其中玩家需要选择一个整数以获得相对于另一名玩家的最大胜利概率。通过分析可知，最佳的选择通常是比对手所选数字大一或小一的数，具体取决于这些选择能够覆盖的范围。

One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to n. Let's assume that Misha chose number m, 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 to c. The boys agreed that if m and a are located on the same distance from c, Misha wins.

Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number n. 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 $\text{[math]}$ is maximal, where c is the equiprobably chosen integer from 1 to n (inclusive).

Input

The first line contains two integers n and m (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.

Examples

Input

3 1

Output

Input

4 3

Output

Note

In the first sample test: Andrew wins if c is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses a = 3, the probability of winning will be 1 / 3. If a = 1, the probability of winning is 0.

In the second sample test: Andrew wins if c is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of a the probability of winning is less.

题目大意：

有一个随机区间【1，n】；已知现在M选择的数字是m.现在让你选择一个数字c，使得你可以获胜的几率最大。每轮游戏系统都会随机出来一个数字，对应获胜的人就是距离这个数字最近的人，如果两个数字距离这个随机的数字是一样近的，M获胜。

思路：

想要获胜，显然我们需要压制住M所选的数字，让随机出来的一片数字距离自己都是最近的才行。

那么很显然，要么选择m-1，要么选择m+1.

进行区间大小的比对即可。。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        if(n==1&&m==1)
        {
            printf("1\n");
            continue;
        }
        int b=m-1;
        int c=m+1;
        if(b>=n-c+1)
        {
            printf("%d\n",b);
        }
        else printf("%d\n",c);
    }
}