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.*;
import java.io.*;

class Main {

	void solve() {

		int a = inint(), b = inint();
		out.println(b>=a ? a : a-1);

	}

	public static Scanner in = new Scanner(System.in);
	public static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) {
		new Main().solve();
		out.flush();
	}

	public int inint() {
		return in.nextInt();
	}

	public String instr() {
		return in.next();
	}

	public int[] inintar(int num) {
		int[] a = new int[num];
		for (int i=0; i<num; i++) {
			a[i] = in.nextInt();
		}
		return a;
	}

	public char[] incharar() {
		return in.next().toCharArray();
	}

}