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.

题目意思：对于t组样例，每一组一共有n个音符，m个音符属于一种，剩下的音符都是不同种的，出现连续同一种音符的个数作为得分，求最大得分和最小得分。

解题思路：其实我们可以这样想一共会有x=n-m+1种音符，最大得分必然是相同音符的数量也就是m，而对于最小得分：如果音符的种类x大于等于同一种的数量m,

那么同一种音符便可以被不同种的音符一一分割开；如果音符的种类小于同一种的数量m,那么想要得到最小连续同一类的音符数，可以理解为将m利用不同种类的音符划分为若干部分。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 int main()
 6 {
 7     int t,a,b,ans,x;
 8     scanf("%d",&t);
 9     while(t--)
10     {
11         scanf("%d%d",&a,&b);
12         x=a-b+1;
13         if(b==0)
14         {
15             ans=0;
16         }
17         else if(x>=b)
18         {
19             ans=1;
20         }
21         else if(x<b)
22         {
23             if(b%x!=0)
24             {
25                 ans=b/x+1;
26             }
27             else
28             {
29                 ans=b/x;
30             }
31         }
32         printf("%d %d\n",b,ans);
33     }
34     return 0;
35 }