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)

(* エラトステネスのふるい *)
let sieve :
  (* 素数かどうかを判定する自然数の上界 *)
  int ->
  (* 素数ならtrueを返す関数 *)
  (int -> bool)
  = fun n ->
    let a = Array.make n true in
    a.(0) <- false;
    a.(1) <- false;
    for p = 2 to int_of_float @@ ceil @@ sqrt @@ float_of_int n do
      if a.(p) then
        for i = p to (n - 1) / p do
          a.(p * i) <- false
        done
    done;
    Array.get a
 
let rec take n = function
  | [] -> []
  | x :: xs ->
      if n = 0 then []
      else x :: take (n - 1) xs
 
let () = Scanf.scanf "%d\n" @@ fun n ->
  let p = sieve 55555 in
  Array.init 55555 (fun i -> i)
  |> Array.to_list
  (*
   * (5i + 1) + (5j + 1) + (5k + 1) + (5l + 1) + (5m + 1) 
   * = 5 (i + j + k + l + m + 1)
   *)
  |> List.filter (fun n -> n mod 5 = 1 && p n)
  |> take n
  |> List.iter (Printf.printf "%d ")