We have the following matrix 
1 0 0 0 0 0 ...
2 2 0 0 0 0 ...
3 3 3 0 0 0 ...
4 4 4 4 0 0 ... 
5 5 5 5 5 0 ... 
6 6 6 6 6 6 ...
and so on ... 

The matrix is created as follows, first row contains one 1 and rest 0's, second row contains 2 twos and rest zeros, third row contains 3 threes and so on.

Given R and C, calculate the count of even and odd numbers in sub matrix [R,C] .
0 is neither odd nor even and 1 based indexing is used i.e. matrix [1,1]=1 

Input:
First line contains number of test cases T.Each test case contains two integers R and C.

Output
For each test case print count of even and odd numbers in sub matrix [R,C].

Constraints:
1 ≤ T ≤ 100000  
1 ≤ R ≤ 100000 
1 ≤ C ≤ 100000  

SAMPLE INPUT
3
2 3
6 7
10 10

SAMPLE OUTPUT
2 1
12 9
30 25
#Even Odd
import sys

t=int(sys.stdin.readline())

while t>0:
	r,c=map(int,sys.stdin.readline().split())

	even=0
	odd=0
	if r<=c:
		if r%2!=0:
			odd=(((r+1)/2)*(1+r))/2
			even=((r/2)*(1+r))/2
		else:
			odd=((r/2)*(r))/2
			even=((r/2)*(2+r))/2
		
	else:
		if c%2!=0:
			odd=(((c+1)/2)*(1+c))/2
			even=((c/2)*(1+c))/2
			if (r-c)%2:
				even+=(((r-c+1)/2)*c)
				odd+=(((r-c)/2)*c)
			
			else:
				even+=(((r-c)/2)*c)
				odd+=(((r-c)/2)*c)
				
		else:
				odd=((c/2)*(c))/2
				even=((c/2)*(2+c))/2
				if (r-c)%2:
					even+=(((r-c)/2)*c)
					odd+=(((r-c+1)/2)*c)
				
				else:
					even+=(((r-c)/2)*c)
					odd+=(((r-c)/2)*c)
				
	print ("{0} {1}".format(even,odd))
	t-=1