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 (Clang 3.8.0) (judge verdict: Accepted)

#include<cstdio>
using namespace std;
int main(){
    int mon[13][31];
    int sum=0;
    for(int i=1;i<=12;i++)
        for(int j=1;j<=31;j++)
        {
            if((i<=7&&i%2==0)&&j==31)
                break;
            if(i==2&&j==29)
                break;
            if(i>=8&&i%2==1&&j==31)
                break;
            if(i==j)
                sum++;
            mon[i][j]=sum;
        }
    int a,b;
    scanf("%d%d",&a,&b);
    {
        printf("%d\n",mon[a][b]);
    }
    return 0;
}