2018 ACM-ICPC 亚洲青岛区域网络赛 A Live Love

DreamGrid is playing the music game Live Love. He has just finished a song consisting of n notes and got a result sequence A​1​​,A​2​​,...,A​n​​ (A​i​​∈ {PERFECT, NON-PERFECT}). The score of the song is equal to the max-combo of the result sequence, which is defined as the maximum number of continuous PERFECTs in the sequence.

Formally speaking, max-combo(A)=max { k | k is an integer and there exists an integer i (1≤i≤n−k+1) such that A​i​​=A​i+1​​=A​i+2​​=...=A​i+k−1​​= PERFECT }. For completeness, we define max(∅)=0.

As DreamGrid is forgetful, he forgets the result sequence immediately after finishing the song. All he knows is the sequence length n and the total number of PERFECTs in the sequence, indicated by m. Any possible score s he may get must satisfy that there exists a sequence A​′​​of length n containing exactly m PERFECTs and (n−m) NON-PERFECTs and max-combo(A​′​​)=s. Now he needs your help to find the maximum and minimum s among all possible scores.

Input

There are multiple test cases. The first line of the input contains an integer T (1≤T≤100), indicating the number of test cases. For each test case:

The only line contains two integers n and m (1≤n≤10​3​​, 0≤m≤10​3​​, m≤n), indicating the sequence length and the number of PERFECTs DreamGrid gets.

Output

For each test case output one line containing two integers s​max​​ and s​min​​, indicating the maximum and minimum possible score.

Sample Input

5
5 4
100 50
252 52
3 0
10 10

Sample Output

4 2
50 1
52 1
0 0
10 10

Hint

Let's indicate a PERFECT as P and a NON-PERFECT as N.

For the first sample test case, the sequence (P,P,P,P,N) leads to the maximum score and the sequence (P,P,N,P,P) leads to the minimum score.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		int n,m;
		cin>>n>>m;
		if(n==m)
		   cout<<m<<" "<<m<<endl;
		else if(m==0)
		   cout<<"0"<<" "<<"0"<<endl;
		else 
		{
			if(n/m>=2)
              cout<<m<<" "<<"1"<<endl;
            else 
            {
            	int c = n-m;
            	int s;
            	if(m%(c+1)==0)
            	   s = m/(c+1);
            	else 
            	   s = m/(c+1) + 1;
            	cout<<m<<" "<<s<<endl;
            }
		}
	}
	return 0;
}