Codeforces Round #316 (Div. 2)570B Simple Game(脑洞)

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 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).

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.

Sample test(s)

input

3 1

output

2

input

4 3

output

2

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.

注意wa点：1 1 输出1

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "map"
using namespace std;
int n, m;
int main(int argc, char const *argv[])
{
	scanf("%d%d", &n, &m);
	int r1 = -1, r2 = -1;
	if(n == 1) printf("1\n");
	else {
		if(m != n) r1 = n - m;
		if(m != 1) r2 = m - 1;
		if(r1 > r2) printf("%d\n", m + 1);
		else printf("%d\n", m -1);
	}
	return 0;
}