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>
typedef long long ll;
typedef long double ld;
const int INF = 1e9, MOD = 1e9 + 7, ohara = 1e6 + 10;
const ll LINF = 1e18;
using namespace std;

#define rep(i, n) for (int(i) = 0; (i) < (int)(n); (i)++)
#define rrep(i, a, b) for (int i = (a); i < (b); i++)
#define rrrep(i, a, b) for (int i = (a); i >= (b); i--)
#define all(v) (v).begin(), (v).end()
#define Size(n) (n).size()
#define Cout(x) cout << (x) << endl

ll n, cnt, ans, a, b, c, d, tmp, tmpp, m, h, w, x, y, sum, pos, k;
ld doua;
int dy[] = {1, 0, -1, 0};
int dx[] = {0, 1, 0, -1};
//int dy[]={-1,0,1,-1,1,-1,0,1};
//int dx[]={-1,-1,-1,0,0,1,1,1};
string alph("abcdefghijklmnopqrstuvwxyz"), s;
bool fl;
struct edge
{
    int to, cost;
};

//-------------------------↓↓↓↓↓↓------------------------

int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> n >> a >> b;
    cout << min(a, b) << " " << max(0LL, a + b - n) << "\n";
    cerr << "Heloo\n";
    return 0;
}