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 Ruby (2.3.3) (judge verdict: Runtime Error)

h,w = gets.split.map(&:to_i)
s = []
h.times do |i|
  s[i] = gets.chomp.split("")
end

h.times do |i|
  w.times do |j|
    if s[i][j] == "#"
      dec = false
      [[i-1,j],[i+1,j],[i,j-1],[i,j+1]].each do |hi,wi|
        next if hi < 0 || hi > h - 1 || wi < 0 || wi > w - 1
        if s[hi][wi] == "#"
          dec = true
          break
        end
      end
      if !dec
        puts "No"
        exit
      end
    end
  end
end

puts "Yes"