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;

#define lint long long
#define P pair<int, int>
#define LLP pair<long long, long long>
#define REP(i, x, n) for(int i = (x), i##_len = (int)(n) ; i < i##_len ; ++i)
#define rep(i, n) for(int i = 0, i##_len = (int)(n) ; i < i##_len ; ++i)
#define repr(i, n) for(int i = (int)(n) - 1 ; i >= 0 ; --i)
#define SORT(x) sort((x).begin(), (x).end())
#define SORT_INV(x) sort((x).rbegin(), (x).rend())

const int IINF = 1e9 + 10;
const long long LLINF = (long long)1e18 + 10;
const long long MOD = (long long)1e9 + 7;
const int dx4[] = {1, 0, -1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};
const double EPS = 1e-8;

int main(){

    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, a, b;
    cin >> n >> a >> b;

    int ans1 = min(a, b), ans2 = max(0, a + b - n);

    cout << ans1 << " " << ans2 << endl;

    return 0;

}