Rnadom Teams

　　Rnadom  Teams

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.actioncid=88890#problem/B

题目：

 Description

n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.

Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.

Input

The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 10⁹) — the number of participants and the number of teams respectively.

Output

The only line of the output should contain two integers k_min and k_max — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.

Sample Input

Input

5 1

Output

10 10

Input

3 2

Output

1 1

Input

6 3

Output

3 6

Hint

In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.

In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.

In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.

分析：

    最大组队数肯定有一队为n-m+1人，最小的为把人数尽可能均分。

即有n%m队为n/m+1,n-n%m队为n/m人，最后算一下朋友对数，

一个队友n人，可以产生n*(n-1)/2队朋友。

#include<iostream>
using namespace std;
int main()
{
  long long mi,n,m,ma,x,i=0;
    cin>>n>>m;
       x=n/m;
    if(n%m!=0)
      i=n%m;
    mi=(x*(x-1)/2)*(m-i)+(x*(x+1)/2)*i;
    ma=((n-m+1)*(n-m)/2);
     cout<<mi<<' '<<ma<<endl;
    return 0;
}