Codeforces 655B B. Mischievous Mess makers (greedy)

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B. Mischievous Mess Makers

Time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

It is a balmy spring afternoon, and Farmer John ' S  n  cows be ruminating Abou T link-cut cacti in their stalls. The cows, Labeled 1 through N , is arranged so The  i -th cow occupies the  i -th stall from The left. However, Elsie, after realizing that she'll forever live in the shadows beyond Bessie ' s limelight, have formed the Mischi Evous 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, distinct stalls and swap the cows occupying those stalls, Making no more than One swap each minute.

Being the meticulous pranksters that they is, the mischievous Mess makers would like to know the maximum messiness attain Able in the K minutes that they has. We denote as pi the label of the cow in the I-th stall. The messiness of an arrangement of cows are defined as the number of pairs (i, J) such that c12>I < J and pi > pJ.

Input

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

Output

Output A single integer, the maximum messiness this mischievous Mess makers can achieve by performing no more than K swaps.

Examplesinput

5 2

Output

10

Input

1 10

Output

0

Note

In the first sample, the mischievous Mess makers can swaps the cows in the stalls 1 and 5 during the first minute, Then the cows in stalls2 and 4 during the second minute. This reverses the arrangement of cows, giving us a total messiness of ten.

In the second sample, there are only one cow, so the maximum possible messiness is 0.

Test instructions: Ask for a maximum of K this interchange, the number of pairs (i, J) such that I < J and p i > pJ.

Idea: From the greedy know, each interchange the most out of a pair can get the largest number, each change the outermost pair, number increase 2*n-3;

AC Code:

#include <bits/stdc++.h>using namespacestd;intMain () {Long Longn,k,ans=0; CIN>>n>>K;  while(k--&&n>1) {ans+=2*n-3; N-=2; } cout<<ans<<Endl; return 0;}

Codeforces 655B B. Mischievous Mess makers (greedy)