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 (1.15.1) (judge verdict: Accepted)

#[allow(dead_code)]
fn read<T: std::str::FromStr>() -> T
{
    std::io::Read::bytes(std::io::stdin())
                    .map(|c| c.unwrap() as char)
                    .skip_while(|c| c.is_whitespace())
                    .take_while(|c| !c.is_whitespace())
                    .collect::<String>()
                    .parse().ok().unwrap()
}

fn main() {
    let a: i32 = read();
    let b: i32 = read();
    let ans: i32;

    if a <= b {
        ans = a;
    } else {
        ans = a - 1;
    }

    println!("{}", ans);
}