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: Runtime Error)

import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);		// 整数の入力


int A= sc.nextInt();
int B= sc.nextInt();
int C= sc.nextInt();
int K= sc.nextInt();
int max=A;
int sonota=B+C;
int sum=0;

	if(max<B && max>C){
		max=B;
		sonota=A+C;
	}else{
		max=C;
		sonota=A+B;
	}

	for(int i=0; i<K; i++){
		sum=max*2+sonota;
		max=2*max;
	}
	System.out.println(sum);
	}

}