Positive Negative Sign

Positive Negative Sign Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit  Status  Practice  LightOJ 1294  uDebug

Description

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 mintegers 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.

Sample Input

2

12 3

4 1

Sample Output

Case 1: 18

Case 2: 2

Hint

Find formula

#include<cstdio> #include<algorithm> using namespace std; int main() { int t,cont=0; long long n,m; scanf("%d",&t); while(t--) { scanf("%lld %lld",&n,&m); printf("Case %d: %lld\n",++cont,n/2*m); } return 0; }