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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void)
{
	int N;
	int t=1;
	int A[100],B[100];
	int i,g;
	scanf("%d",&N);
	//printf("%d\n",N);
	int Amax=0,Bmax=0;
	int AP=0,BP=0;
	
	for(i=0;i<N;i++){
	scanf("%d%d",&A[i],&B[i]);
	}
	
	for(g=0;g<N;g++){
	    if(t==1){
	for(i=0;i<N;i++){
	if(A[Amax]<=A[i]){
		Amax=i;
		B[i]=0;
		AP+=A[Amax];
		//printf("a %d\n",A[Amax]);
		A[Amax]=0;
		break;
		}
	}
	        t=0;
	    }
	else{
	for(i=0;i<N;i++){
	if(B[Bmax]<=B[i]){
		Bmax=i;
		A[i]=0;
		BP+=B[Bmax];
		//printf("b %d\n",B[Bmax]);
		B[Bmax]=0;
		break;
		}
	}t=1;
	}
	}
	printf("%d",AP-BP);
}