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: Runtime Error)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws NumberFormatException,
			IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String str;
		while ((str = br.readLine()) != null) {
			String inputs[] = str.split(" ");
			int a = Integer.parseInt(inputs[0]);
			int b = Integer.parseInt(inputs[1]);
			int c = Integer.parseInt(inputs[2]);
			int k = Integer.parseInt(br.readLine());
			for (int i = 0; i < k; i++) {
				if (a > b && a>c) {
					a *= 2;
				} else if (b > c) {
					b *= 2;
				} else {
					c *= 2;
				}
			}
			System.out.println(a+b+c);
		}
	}
}