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;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using vi = vector<int>;
using vb = vector<bool>;
using vvi = vector<vector<int>>;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using pii = pair<int,int>;
#define FOR(i,a,b) for(ll i = (a); i < (ll)(b); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(obj) (obj).begin(), (obj).end()
#define eb(val) emplace_back(val)
const double PI = acos(-1);
const double EPS = 1e-10;
const ll MOD = 1e9+7;
void cioacc(){//accelerate cin/cout
  cin.tie(0);
  ios::sync_with_stdio(false);
}
signed main(){
  int n,a,b;
  cin >> n >> a >> b;
  int ans1 = min(a,b);
  int ans2 = max(a+b-n,0);
  cout << ans1 << " " << ans2 <<endl;
}