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 OCaml (4.02.3) (judge verdict: Runtime Error)

let main () =
  let a, b, c = Scanf.scanf "%d %d %d\n" (fun a b c -> a, b, c) in
  let k = Scanf.scanf "%d\n" (fun a -> a) in
  let rec pow i j = if j = 1 then i else pow (i * i) (j - 1) in
  let ans1 = a * (pow 2 k) + b + c in
  let ans2 = a + b * (pow 2 k) + c in
  let ans3 = a + b + c * (pow 2 k) in
  Printf.printf "%d\n" (max (max ans1 ans2) ans3);;

let _ = main ()