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>

#define null               ""
#define all(dat)           dat.begin(), dat.end()
#define over(msg)          cout << msg << endl, exit(0);
#define loop(i, to)        for (int i = 0; i < to; ++i)
#define cont(i, to)        for (int i = 1; i <= to; ++i)
#define circ(i, fr, to)    for (int i = fr; i <= to; ++i)
#define foreach(i, dat)    for (__typeof(dat.begin()) i = dat.begin(); i != dat.end(); ++i)

typedef long long          num;

using namespace std;

int n, a, b;

int main() {
    ios_base::sync_with_stdio(0);
    cin >> n >> a >> b;
    cout << min(a, b) << ' ' << max(a + b - n, 0) << endl;
}