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 C# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
class Hello{
	public static void Main(){
		string input = Console.ReadLine();
		string inputtimes = Console.ReadLine();
		int times = Int32.Parse(inputtimes);
		string[] split = input.Split(' ');
		int[] array = new int [split.Length];
		int max = 0;
		int sum = 0;
		for(int count = 0;count < split.Length;count++){
			array[count] = Int32.Parse(split[count]);
			sum = sum + array[count];
			if(array[count] > max){
				max = array[count];
			}
		}
		int exmax = max;
		for(int i = 0;i < times;i++){
			max = max * 2;
		}
		int result = max + sum - exmax;
		Console.WriteLine(result);
	}
}