題解：Mischievous Mess Makers

time limit per test

 1 second

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

It is a balmy spring afternoon, and Farmer John's n cows are ruminating about link-cut cacti in their stalls. The cows, labeled 1 throughn, are arranged so that the i-th cow occupies the i-th stall from the left. However, Elsie, after realizing that she will forever live in the shadows beyond Bessie's limelight, has formed the Mischievous Mess Makers and is plotting to disrupt this beautiful pastoral rhythm. While Farmer John takes his k minute long nap, Elsie and the Mess Makers plan to repeatedly choose two distinct stalls and swap the cows occupying those stalls, making no more than one swap each minute.

Being the meticulous pranksters that they are, the Mischievous Mess Makers would like to know the maximum messiness attainable in the k minutes that they have. We denote as pi the label of the cow in the i-th stall. The messiness of an arrangement of cows is defined as the number of pairs (i, j) such that i < j and pi > pj.

Input

The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of cows and the length of Farmer John's nap, respectively.

Output

Output a single integer, the maximum messiness that the Mischievous Mess Makers can achieve by performing no more than k swaps.

Examples

input

5 2

output

10

input

1 10

output

0

分析：

使用貪心算法。很容易驗證最佳算法爲每次交換頭尾兩個數字。k>n/2時，達到極限。

Code:

#include<stdio.h>

int main()
{
	long long int n,k;
	
	long long int i,j;
	long long int count = 0 ;
	
	int temp;
	scanf("%lld%lld",&n,&k);
	
	
	if(k>(n/2))
		{printf("%lld",n*(n-1)/2);
		 return 0;
		}
		
	count=(2*n-k-1)*k/2+(k-1)*k/2+(n-2*k)*k;
	printf("%lld",count);
	return 0;
}