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 <image> 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

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.
#pyrival orz
import os
import sys
import math
from io import BytesIO, IOBase


input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))

############ ---- Dijkstra with path ---- ############
def dijkstra(start, distance, path, n):
    # requires n == number of vertices in graph,
    # adj == adjacency list with weight of graph

    visited = [False for _ in range(n)]  # To keep track of vertices that are visited
    distance[start] = 0  # distance of start node from itself is 0

    for i in range(n):
        v = -1  # Initialize v == vertex from which its neighboring vertices' distance will be calculated
        for j in range(n):
            # If it has not been visited and has the lowest distance from start
            if not visited[v] and (v == -1 or distance[j] < distance[v]):
                v = j

        if distance[v] == math.inf:
            break

        visited[v] = True  # Mark as visited

        for edge in adj[v]:
            destination = edge[0]  # Neighbor of the vertex
            weight = edge[1]  # Its corresponding weight

            if distance[v] + weight < distance[destination]:  # If its distance is less than the stored distance
                distance[destination] = distance[v] + weight  # Update the distance
                path[destination] = v  # Update the path
        

def gcd(a, b):
    if b == 0:
        return a
    else:
        return gcd(b, a%b)

def lcm(a, b):
    return (a*b)//gcd(a, b)

def ncr(n, r):
    return math.factorial(n)//(math.factorial(n-r)*math.factorial(r))

def npr(n, r):
    return math.factorial(n)//math.factorial(n-r)

def seive(n):
    primes = [True]*(n+1)
    ans = []

    for i in range(2, n):
        if not primes[i]:
            continue

        j = 2*i
        while j <= n:
            primes[j] = False
            j += i

    for p in range(2, n+1):
        if primes[p]:
            ans += [p]

    return ans

def factors(n):
    factors = []

    x = 1
    while x*x <= n:

        if n % x == 0:
            if n // x == x:
                factors.append(x)
            else:
                factors.append(x)
                factors.append(n//x)

        x += 1

    return factors

# Functions: list of factors, seive of primes, gcd of two numbers,
# lcm of two numbers, npr, ncr

def main():
    try:
        n, m = invr()
        if n == 1:
            print(n)

        elif m - 1 >= n - m:
            print(m-1)
        else:
            print(m+1)

    except Exception as e:
        print(e)

        

# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":
    main()