Problem:
Score: 100 points

Problem Statement

In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year.

For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year.

In this country, a date is called Takahashi when the month and the day are equal as numbers. For example, 5-5 is Takahashi.

How many days from 2018-1-1 through 2018-a-b are Takahashi?

Constraints

a is an integer between 1 and 12 (inclusive).

b is an integer between 1 and 31 (inclusive).

2018-a-b is a valid date in Gregorian calendar.

Input

Input is given from Standard Input in the following format:

a b

Output

Print the number of days from 2018-1-1 through 2018-a-b that are Takahashi.

Sample Input 1

5 5

Sample Output 1

5

There are five days that are Takahashi: 1-1, 2-2, 3-3, 4-4 and 5-5.

Sample Input 2

2 1

Sample Output 2

1

There is only one day that is Takahashi: 1-1.

Sample Input 3

11 30

Sample Output 3

11

There are eleven days that are Takahashi: 1-1, 2-2, 3-3, 4-4, 5-5, 6-6, 7-7, 8-8, 9-9, 10-10 and 11-11.

Solution in Haskell (GHC 7.10.3) (judge verdict: Runtime Error)

{-# LANGUAGE ScopedTypeVariables #-}

import Data.Array
import Data.List
import Control.Monad

solve arr h w
    | null res = "Yes"
    | otherwise = "No"
    where
        res = [() | i <- [1..h], j <- [1..w], step arr i j]
        step arr i j = curr && ns
            where
                curr = arr ! (i,j) == '#'
                ns = null $ (filter (\p -> arr ! p == '#') $ filter (\(a,b) -> a > 0 && a <= h && b > 0 && b <= w) [(i+1,j),(i-1,j),(i,j+1),(i,j-1)])


main = do
    [h, w] <- map read . words <$> getLine :: IO [Int]
    str <- concat <$> replicateM h getLine :: IO String
    let (grid :: Array (Int,Int) Char) = listArray ((1,1),(h,w)) str
    putStrLn $ solve grid h w