Maxim Buys an Apartment
Maxim wants to buy an apartment in a new house at Line Avenue of Metropolis. The house has n apartments that are numbered from 1 to 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 and 5 were inhabited. In this case all other apartments: 2, 4 and 6 are good.
题意：n个房间，有m个有人住了，但是不知道是那一个。有人住的房间的左边和右边都可以买。求最少和最多能卖几个。
思路：如果m<n的话,最少是1.
#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
int n,k;
cin>>n>>k;
int ans=0;
if(n==k||k==0)
cout<<0<<" "<<0<<endl;
else
{
cout<<1<<" "<<min(n-k,k*2)<<endl;
}
}