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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"

using namespace std;
using ll=long long;
using ull=unsigned long long;

typedef pair<int,int>Pi;
typedef pair<long long,long long>P;
typedef pair<long long,P>PP;
typedef pair<P,P>PPP;

const long long MOD=1e9+7;
const long long INF=5e18;
const int di[4]={1,0,-1,0};
const int dj[4]={0,1,0,-1};

#define fr first
#define sc second
#define pb push_back
#define eb emplace_back
#define ALL(x) (x).begin(),(x).end()
#define ALLR(x) (x).rbegin(),(x).rend()

int a,b,n;

void input(){
	cin>>n>>a>>b;
	return;
}

void solve(){
	cout<<min(a,b)<<' '<<max(0,min(a,b)-(n-(max(a,b))))<<endl;
	return;
}

int main(){
	input();
	solve();
	return 0;
}