cf 570B Simple Game 水题

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B. Simple Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

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.

题意：给定n和m，找到一个数a，使几率最大

思路：a在1-n之间，m已知，使a到c的距离小于m到c的距离

如果2m>n，则m位于靠近n的一侧，此时只要取<m的数即可，同理，2m<n，则m位于靠近1的一侧，此时只要取>m的数即可

水题，但还是要写，因为自己绝对值这方面不怎么理解，比赛时也是误打误撞写出来的

代码：

#include<stdlib.h>#include<stdio.h>#include<cmath>#include<algorithm>#include<string>#include<string.h>#include<set>#include<queue>#include<stack>#include<vector>#include<functional> #include<map>using namespace std;const int maxn = 200 + 10;const int INF = (int)1e9;int n, m;int c, a;////判断|c-a|<|c-m|int main(){while (scanf("%d %d", &n, &m) != EOF) {if (n == 1)printf("1\n");//!!!else {if (n >= m * 2)printf("%d\n", m + 1);else printf("%d\n", m - 1);}}//system("pause");return 0;}