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;

typedef long long ll;
typedef pair<int, int> pii;

#define rep(i, n) for (int i = 0; i < (n); i++)
#define repp(i, s, e) for (int i = (s); i < (e); i++)
#define all(x) x.begin(), x.end()

#define endl "\n"

void use_cio() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
}

template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
    return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
    os << "[";
    for (auto x : V) os << x << ", ";
    return os << "]";
}
//--------------------------------------------------//

int main() {
    use_cio();
    int n, a, b;
    cin >> n >> a >> b;
    cout << min(a, b) << " " << max(0, a + b - n);
}