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 Common Lisp (SBCL 1.1.14) (judge verdict: Wrong Answer)

(defun sieve (n)
  (let ((is-prime (make-array (1+ n)
                              :initial-element t))
        (primes '()))
    (setf (svref is-prime 0) nil)
    (setf (svref is-prime 1) nil)
    (loop for i from 2 to n
          do (when (svref is-prime i)
               (push i primes)
               (loop for j from (* 2 i) to n by i
                     do (setf (svref is-prime j) nil))))
    (nreverse primes)))

(defparameter goods
  (remove-if (lambda (n) (/= (mod n 5) 1))
             (sieve 55555)))

(defparameter n (read))
(format t "~{~a~^ ~}~%"
        (subseq goods 0 n))