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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(unused_imports)]
#![allow(dead_code)]
use proconio::input;
use proconio::fastout;
use proconio::marker::{Chars};

#[fastout]
fn main() {
    input! {
        a:i32,b:i32,
    }
    let v:[i32;12]=[31,28,31,30,31,30,31,31,30,31,30,31];
    let mut ans=0;
    for i in 1..a+1{
        if i==a{
            for j in 1..b+1{
                if i==j{
                    ans+=1;
                }
            }
        }else{
            for j in 1..v[(i-1) as usize]+1{
                if i==j{
                    ans+=1
                }
            }
        }
    }
    println!("{}",ans);
}