2015 HUAS Summer Contest#5~B

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.

解题思路：题目的意思是n个人参加比赛分成m个团队，比完赛之后来自于同一团队的每个人都互相成为了朋友，问能够成为最大和最小数量的朋友分别是多少。首先由于数量的问题，变量应该是long long 型。求最大数量其实就是让其它的团队的人数都为1，然后让剩下的都成为最后一组的团岁的人数，这样就可以使形成的朋友数量最大；求最小数量就是先将人数除以需要分配的组数，然后再将余下的数分配给其它的组，这样题目就可以过了。

程序代码：

#include<stdio.h>
long long t,n,m,v,w,i,kmin,kmax;
int main()
{

    scanf("%lld%lld",&n,&m);
    t=n;
    for(i=1;i<=m;i++)
        t--;
    kmax=((1+t)*t)/2;
    v=n/m;
    w=v*m;
    if(w==n)
    {
        kmin=m*((v-1)*v)/2;
    }
    else
    {
        int s=n-w;
        kmin=(m-s)*((v-1)*v)/2+s*((1+v)*v)/2;
    }
    printf("%I64d %I64d
",kmin,kmax);
    return 0;
}