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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

namespace NKI2019A
{
    static class MainClass
    {

        static int[] s;
        static int[] n;

        public static void Main(string[] args)
        {

            s = Array.ConvertAll(Console.ReadLine().Split().ToArray(), int.Parse);


            Console.WriteLine(Math.Min(s[1], s[2]) + " " + Math.Max(s[1] + s[2] - s[0], 0));

            return;


        }



        public static long Gcd(long a, long b)
        {
            long tmp;
            // a<bにする
            if (a >= b)
            {
                tmp = a;
                a = b;
                b = tmp;
            }

            if (a == 0)
            {
                return b;
            }

            tmp = a;
            a = b % a;
            b = tmp;
            return Gcd(a, b);

        }


        public static IEnumerable<long> PF(long n)
        {
            long i = 2;
            long tmp = n;

            while (i * i <= n) //※1
            {
                if (tmp % i == 0)
                {
                    tmp /= i;
                    yield return i;
                }
                else
                {
                    i++;
                }
            }
            if (tmp != 1) yield return tmp;//最後の素数も返す
        }




    }

}