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<iostream>
#include <algorithm>
#include <vector>
using namespace std;

int N;

typedef long long ll;

ll T = 0,AO = 0,ans = 0;

struct Data {
  ll  A,B,C;
};

int main() {
  cin >> N;
  vector<Data> data(N);
  for ( int i = 0; i < N; i++ ) {
    cin >> data[i].A >> data[i].B;
    data[i].C = data[i].A + data[i].B;
  }

   sort(data.begin(),data.end(),[](const Data &l, const Data&r){
  				 return (l.C > r.C); });
 
  
 bool f = true;
 for ( int i = 0; i < N; i++ ) {
    if (f)  {
      T += data[i].A;
       f = false;
    }
    else {
      AO += data[i].B;
       f = true;
    }
  }
  ans = T - AO;
    cout << ans << endl;
  
  return 0;
}