​​http://www.elijahqi.win/archives/720​​​
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1to n and are arranged in a row. Two apartments are adjacent if their indices differ by 1. Some of the apartments can already be inhabited, others are available for sale.

Maxim often visits his neighbors, so apartment is good for him if it is available for sale and there is at least one already inhabited apartment adjacent to it. Maxim knows that there are exactly k already inhabited apartments, but he doesn’t know their indices yet.

Find out what could be the minimum possible and the maximum possible number of apartments that are good for Maxim.

Input
The only line of the input contains two integers: n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ n).

Output
Print the minimum possible and the maximum possible number of apartments good for Maxim.

Example
input
6 3
output
1 3
Note
In the sample test, the number of good apartments could be minimum possible if, for example, apartments with indices 1, 2 and 3 were inhabited. In this case only apartment 4 is good. The maximum possible number could be, for example, if apartments with indices 1, 3 and5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.

这里有几个坑点，k*3根据题目给的数据范围可能会超过int

还有就是k==0的时候特判一下

#include<cstdio>
int n,k;
int main(){
    //freopen("cfb.in","r",stdin);
    scanf("%d%d",&n,&k);
    if (n==k||k==0){
        printf("0 0");return 0;
    }
    long long tmp=(long long)k*3;
    if (n>=tmp) {printf("%d %d",1,k<<1);return 0;}
    printf("%d %d",1,n-k);
    return 0;
}