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: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Console;
using static System.Math;
 
public class Hello{
    public static void Main(){
        string[] input = ReadLine().Split(' ');
        int n = int.Parse(input[0]);
        int m = int.Parse(input[1]);
//        int a = int.Parse(input[2]);
//        int b = int.Parse(input[3]);*/
//        string temp = ReadLine();
//        int a = int.Parse(ReadLine());
//        int b = int.Parse(ReadLine());
//        int c = int.Parse(ReadLine());
//        int d = int.Parse(ReadLine());
//        string a = ReadLine();
//        string b = ReadLine();
//        string c = ReadLine();
        if(n > m){
            WriteLine(n-1);
        }else{
            WriteLine(n);
        }
    }
}