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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
int main(void){
    int h,w;
    int count;
    char box[60][60];
    int h1=scanf("%d",&h);
    int w1=scanf("%d",&w);
    for(int i=1;i<=h;i++){
        for(int j=1;j<=w;j++){
            scanf("%s",&box[i][j]);  
        //    printf("%d",box[i][j]);
        }
       // printf("\n");

    } for(int i=1;i<=h;i++){
        for(int j=1;j<=w;j++){
            if(box[i][j]=='#'&&box[i+1][j]!='#'&&box[i-1][j]!='#'&&box[i][j+1]!='#'&&box[i][j-1]!='#'){
                count++;
            }
        }
    }        
            if(count==0){
                printf("Yes");
            }else{
                printf("No");
            }
        
     
     return 0;
}