Simple Game

给 n，m (m <= n)，求一个数 a(1<= a <=n), 使得当 c 在 1 到 n 的整数中 随机取值时 ,|c-a| < |c-m| 成立的概率最大。 感谢@凉凉 提供的翻译

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  is maximal, where $ c $ is the equiprobably chosen integer from $ 1 $ to $ n $ (inclusive).

The first line contains two integers $ n $ and $ m $ ( $ 1<=m<=n<=10^{9} $ ) — the range of numbers in the game, and the number selected by Misha respectively.

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.

3 1

2

4 3

2

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.