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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.time.chrono.MinguoChronology;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int a = sc.nextInt();
		int b = sc.nextInt();
		int c = sc.nextInt();
		int ansMin = 0;
		int ansMax = 0;
		if(a == b && b==c) {
			ansMin = a;
			ansMax = a;
		} else if ( a/2 <= b && a/2 <= c && a <= b+c) {
			if(b > c) {
				ansMin = b-a/2;
				ansMax=c;
			} else {
				ansMin = c-a/2;
				ansMax=b;
			}
		}else if (( a/2 <= b || a/2 <= c) && a <= b+c) {
			ansMin = b+c-a;
			if(b > c) {
				ansMax=c;
			} else {
				ansMax=b;
			}
		}else if ( 0 == b || c==0) {
		ansMin = 0;
		ansMax=0;
	}
		else {
			ansMin = 0;
			if(b > c) {
				ansMax=c;
			} else {
				ansMax=b;
			}
		}
		System.out.println(ansMax+" "+ansMin);
		
	}

}