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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>

int main(){
    int i,j,h,w;
    scanf("%d %d",&h,&w);
    char s[h+2][w+2];
    for(i=1;i<=h;i++){
            scanf("%s",s[i]+1);
    }
    for(i=0;i<=h+1;i++){
        s[i][0]='.';
        s[i][w+1]='.';
    }
    for(i=0;i<=w+1;i++){
        s[0][i]='.';
        s[h+1][i]='.';
    }
    for(i=1;i<=h;i++){
        for(j=1;j<=w;j++){
            if(s[i][j]=='#' && s[i-1][j]=='.' && s[i][j-1]=='.' && s[i+1][j]=='.' && s[i][j+1]=='.'){
                printf("No\n");
                return 0;
            }
        }
    }
    printf("Yes\n");
    return 0;
}