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 Java (OpenJDK 1.8.0) (judge verdict: Accepted)



import java.io.IOException;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) throws IOException {

		Scanner scan = new Scanner(System.in);

		int a = scan.nextInt();
		int b = scan.nextInt();

		if (a <= b) {
			System.out.println(a);
		} else {
			System.out.println(a - 1);
		}
	}

	//***********************以下共通関数************************************************
	/**
	 * 文字列を反転して返す
	 * @param pStr 文字列
	 * @return 反転した文字列
	 */
	public static String myReverse(String pStr) {
		if (pStr.length() < 1) {
			return "";
		}

		StringBuffer sb = new StringBuffer(pStr);
		return sb.reverse().toString();
	}

	/**
	 * 余りを求める。
	 * @param a
	 * @param b
	 * @return 余り。余りがマイナスの時はプラスで返す。
	 */
	public static long myMod(long a, long b) {
		long ret = 0;

		ret = a % b;

		if (ret < 0) {
			ret += b;
		}

		return ret;
	}

	/**
	 * 乗数を求める ※mod対応
	 * @param kisu 基数
	 * @param josu 乗数
	 * @param primeNumMod_0 法 mod値 使わない場合は「0」
	 * @return
	 */
	public static long myPowWithMod(long kisu, long josu, long primeNumMod_0) {

		long ans = 1;

		//乗数が0の場合→1
		if (josu == 0)
			return 1;

		if (primeNumMod_0 == 0) {
			//modを使わないとき

			while (josu > 0) {

				if (josu % 2 == 0) {
					//累乗が偶数の時

					kisu *= kisu;
					josu = josu / 2;
				} else {
					//累乗が奇数の時
					ans *= kisu;
					josu--;
				}
			}

			return ans;

		} else {
			//modを使うとき

			//初めに桁を落とす
			kisu = myMod(kisu, primeNumMod_0);

			while (josu > 0) {

				if (josu % 2 == 0) {
					//累乗が偶数の時

					kisu *= myMod(kisu, primeNumMod_0);
					josu = josu / 2;

				} else {
					//累乗が奇数の時
					ans *= myMod(kisu, primeNumMod_0);
					josu--;

				}
			}

			return myMod(ans, primeNumMod_0);

		}

	}

}