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 kmin and kmax — 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.

虽然是道水题，但是我还是想说下。

题意：已知∑ai=n(i从1到m),求∑C(2,ai)的最大值和最小值

要求ai>=1

化简：原式=(∑(ai^2)-n)/2

所以只要求出∑(ai^2)的最值即可

考虑我们分配2个数x，y，根据贪心：

x^2+y^2<=(x-1)^2+(y+1)^2  (x<y)

所以要让平方和更大，我们应该从x中拿更多的给y，最终就是x=1，y尽量大

y再和其他数比较，也是同样的结果，最终，我们得到：

当m-1个为1，1个为n-m+1的时候，得到最大值

同理，当m个数尽量平均的时候，就得到最小值。

#include<cstdio>

#include<iostream>

#define LL long long

using namespace std;

int main()

{

    LL n,m;

    cin>>n>>m;

    LL tmax=,tmin=;

    tmax=m-+(n-m+)*(n-m+);

    LL a=n/m;

    LL b=n%m;

    int i=;

    tmin=b*(a+)*(a+)+(m-b)*(a*a);

    cout<<(tmin-n)/<<" "<<(tmax-n)/<<endl;

    return ;

}