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)

# Your code here!
h, w = gets.split(' ').map(&:to_i)
table = Array.new(h){Array.new(w)}
table.map! { |row|
    row = gets.chomp.split('')
}
checker = Array.new(4)
can_draw = true
for row in 0...h do
    for col in 0...w do
        if table[row][col] == "#" then
            checker[0] = table[row - 1][col] if row > 0
            checker[1] = table[row + 1][col] if row < h - 1
            checker[2] = table[row][col - 1] if col > 0
            checker[3] = table[row][col + 1] if col < w - 1
            can_draw = false if checker.count("#") == 0
            p "行#{row} 列#{col} can_draw=#{can_draw}"
        end
    end
end
puts can_draw ? "Yes" : "No"