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.util.*;
import java.lang.Math;
class Main {
    public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      int H = sc.nextInt();
      int W = sc.nextInt(); sc.nextLine(); //nextIntの後残る改行文字を消す
      String[] line = new String[W];
      int r=0;
      for(int i=0;i<W;i++){
      	line[i]=sc.nextLine();
      } 
      String[][] a = new String[H][W];
      String[][] b = new String[H+2][W+2];
         for(int j=0;j<W;j++){
          for(int k=0;k<H;k++){
           a[k][j]=String.valueOf(line[j].charAt(k));
           b[k+1][j+1]=a[k][j];
          }
         }

         for(int j=0;j<W+2;j++){
          for(int k=0;k<H+2;k++){
           b[0][j]=".";
           b[H+1][j]=".";
           b[k][0]=".";
           b[k][W+1]=".";
          }
         }
          
         for(int j=0;j<W+2;j++){
          for(int k=0;k<H+2;k++){
          if(b[k][j].equals("#")){
                   if(b[k+1][j].equals(".")&&b[k-1][j].equals(".")&&b[k][j+1].equals(".")&&b[k][j-1].equals(".")){
                                       r+=1;
                      }else{
                                       r+=0;
                       }
           }
          }
         }
           if(r>0){
                 System.out.println("No");
           }else{
                 System.out.println("Yes");      
           }
   }
}