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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
      String[] nyu=sc.nextLine().split(" ");

  //   String nyu=sc.nextLine();
int A=Integer.parseInt(nyu[0]);
int B=Integer.parseInt(nyu[1]);
//int C=Integer.parseInt(nyu[2]);
//int D=Integer.parseInt(nyu[3]);


//double A=Double.parseDouble(nyu[0]);
//double B=Double.parseDouble(nyu[1]);
/*
switch(C){
case 7:
case 5:
case 3:
System.out.println("YES") ;
break;
default:
System.out.println("NO") ;

break;
}
*/
int flag=0;
if(A<=B){
flag=1;
}

if(A==1){
System.out.println(flag);
}else{
System.out.println(flag+(A-1));

}




	}
}