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.

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.

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.

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 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.

算法分析：1.  n-1 个队有一个人，其余人都放到地n队，这样组合出来最多。

2.  将n个人尽量均分到每个队，这样获得组合最少。

3.  m==1  和  m==n的情况进行一下剪枝。

#include <iostream>

#include <stdio.h>

#include <math.h>

#include <string.h>

#include <algorithm>

using namespace std;

int main()

{

    long long n, m;

    long long dd, ff, gg, d;

    long long mi, ma;

    while(scanf("%lld %lld", &n, &m)!=EOF) //鉴于此处为什么用long long，我也不清楚，就因为这两处的数据类型，我WA了8遍。

    {

        if(m==1)

        {

            dd=n*(n-1)/2;

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

            continue;

        }

        if(n==m)

        {

            printf("0 0\n");

            continue;

        }

        dd=n-(m-1);

        ma=dd*(dd-1)/2;

        ff=n/m;

        gg=n%m;

        d=n-ff*m;

        mi=ff*(ff-1)/2*(m-d);

        mi=mi+ff*(ff+1)/2*d;

        if(mi>ma)

        {

            mi=mi^ma; ma=ma^mi; mi=mi^ma;

        }

        printf("%lld %lld\n", mi, ma );

    }

    return 0;

}