Problem:
Score : 100 points

Problem Statement

We conducted a survey on newspaper subscriptions.
More specifically, we asked each of the N respondents the following two questions:

Question 1: Are you subscribing to Newspaper X?

Question 2: Are you subscribing to Newspaper Y?

As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2.

What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y?

Write a program to answer this question.

Constraints

1 \leq N \leq 100

0 \leq A \leq N

0 \leq B \leq N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the maximum possible number and the minimum possible number of respondents subscribing to both newspapers, in this order, with a space in between.

Sample Input 1

10 3 5

Sample Output 1

3 0

In this sample, out of the 10 respondents, 3 answered they are subscribing to Newspaper X, and 5 answered they are subscribing to Newspaper Y.

Here, the number of respondents subscribing to both newspapers is at most 3 and at least 0.

Sample Input 2

10 7 5

Sample Output 2

5 2

In this sample, out of the 10 respondents, 7 answered they are subscribing to Newspaper X, and 5 answered they are subscribing to Newspaper Y.

Here, the number of respondents subscribing to both newspapers is at most 5 and at least 2.

Sample Input 3

100 100 100

Sample Output 3

100 100

Solution in Rust (1.15.1) (judge verdict: Accepted)

#![allow(non_snake_case)]
#![allow(unused_variables)]
#![allow(dead_code)]

fn main() {
    let (N, A, B): (i64, i64, i64) = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        let mut iter = line.split_whitespace();
        (
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap()
        )
    };

    let ans = format!("{} {}", std::cmp::min(A, B), std::cmp::max(0, A + B - N));

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