Given two integers: n and m and n is divisible by 2m, you have to write down the first n natural numbers in the following form. At first take first m integers and make their sign negative, then take next m integers and make their sign positive, the next m integers should have negative signs and continue this procedure until all the n integers have been assigned a sign. For example, let n be 12 and m be 3. Then we have

-1 -2 -3 +4 +5 +6 -7 -8 -9 +10 +11 +12

If n = 4 and m = 1, then we have

-1 +2 -3 +4

Now your task is to find the summation of the numbers considering their signs.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing two integers: n and m (2 ≤ n ≤ 10⁹, 1 ≤ m). And you can assume that n is divisible by 2*m.

Output

For each case, print the case number and the summation.

#include<stdio.h>

#include<string.h>

#define LL long long

int main()

{

	int t,k;

	LL n,m,sum;

	scanf("%d",&t);

	k=1;

	while(t--)

	{

		scanf("%lld%lld",&n,&m);

		sum=(n/2)*m;

		printf("Case %d: ",k++);

		printf("%lld\n",sum);

	}

	return 0;

}