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 <iostream>
#include <iomanip>
#include <vector>
#include <string>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <functional>
#include <limits.h>
#include <set>
#include <map>
#include <tuple>
using namespace std;

#define ll long long
#define ull unsigned long long
#define rep(i,N) for (int i=0;i<N;i++)
#define loop(i,N,M) for(int i=N;i<M;i++)
#define MAX(v) *max_element(v.begin(),v.end())
#define MIN(v) *min_element(v.begin(),v.end())
#define SORTL2S(type,v) sort(v.begin(),v.end(),greater<type>())
#define SORTS2L(type,v) sort(v.begin(),v.end())
#define SORTMF(v,func) sort(v.begin(),v.end(),func)

//変数
ll N,A,B;

//解法
void solve() {

	cout << min(A, B) << " ";


	if (A + B >= N) {
		cout << (A + B) - N;
	}
	else {
		cout << 0;
	}

	return;
}

int main() {
	//入力

	cin >> N>>A>>B;

	solve();
	cout << endl;
	return 0;
}