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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        // 自分の得意な言語で
        // Let's チャレンジ！！

        Scanner sc = new Scanner(System.in);
        int a = sc.nextInt();
        int b = sc.nextInt();
        int cnt = 0;
        
        if(!(a>=1 && a<=12)) {
        	cnt = 0;
        }else {
        	if(!(b>=1 && b<=31)) {
        		cnt=0;
        	}else {
        		for(int i=1;i<a+1;i++) {
        			cnt++;
        			if(i==a) {
        				if(!(b-i>=0)) {
        					cnt--;
        				}
        			}
        		}System.out.println(cnt);
        	}
        }
    }
}