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: Wrong Answer)


fn main(){
	let mut s = String::new();
	std::io::stdin().read_line(&mut s).unwrap();
	let vec: Vec<&str> = s.trim().split(' ').collect();
	let nm: i32 = vec[0].parse().unwrap();
	let nd: i32 = vec[1].parse().unwrap();
	let mut c = 0;
// 2, 4, 6, 9, 11
	for m in 1..((nm + 1) as usize) {
		for d in 1..((nd + 1)as usize) {
			if m == 2 || m == 4 || m == 6 || m == 9 || m == 11 {
				if d == 31 {
					break;
				}
				if d == 30 && m == 2 {
					break;
				}
			}
			if m == d {
				c += 1;
			}
		}
	}
	println!("{}", c);
}