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: Wrong Answer)

#include <bits/stdc++.h>
#include <time.h>
using namespace std;

typedef long long int LL;
typedef long long int LLint;

//Faster Program
#define endl "\n"

//Faster Coding
#define ED return
#define GET(a) int a;cin >> a
#define DGET(a) double a;cin >> a
#define SGET(a) string a;cin >> a
#define CGET(a) char a;cin >> a
#define PRINT(a) cout << a << "\n"
#define ANS(a,b) int a;cin >> a;cout << b << "\n"
#define FOR(n) for(int i = 0;i < n;i++)
#define SORT(vec) sort(vec.begin(),vec.end())
#define REV(vec) reverse(vec.begin(),vec.end())

//As Function
#define UP(a,b) ((a+(b-1))/b)
#define ipow(x,y) LL(pow(x,y))

//Debug
#define TEST cout << "OK" << endl

const double PI = M_PI;
const long long INF = int(10e9-1);
const long long MOD = int(10e9+7);

//Settings
const bool DO_BENCHMARK = false;
const int DOUBLE_ACCURACY = 15;

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(DOUBLE_ACCURACY);
    void solve();
    if(DO_BENCHMARK){
        clock_t start = clock();
        solve();
        clock_t end = clock();
        const double time = static_cast<double>(end-start)/CLOCKS_PER_SEC*1000.0;
        cout << "\n" << time << "[ms]\n";
    }
    else solve();
}

void solve(){
   GET(N);
   GET(A);
   GET(B);
   cout << min(A,B) << " " << max(N-A-B,0) << endl;
}