ACM 组队交朋友

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：

这个题目的大意是给出n个人，让我们这n个人分成m个组，其中在一起的一个组的人都成为朋友，问我们要如何分组才能使所得的朋友总数最多和最少，然后输出最多和最少的朋友数目，同时要保证每个队里都要有人。最多时，就是把其中一个队分最多的人数，让其他的队的人数都为1；最少时，也就是需要我们进行平均分配，使所有队的人数最多相隔一个。例如把11个人分成3组是这样分配的朋友总是最少：3 4 4；最多 1 1 9；

程序代码：

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
    long long n,m,i,t,x,max,min,r;
    scanf("%lld%lld",&n,&m);
    x=n-m;
    max=((1+x)*x)/2;//等差数列求和公式
    t=n/m;
    r=n%m;
    if(!n)
        min=m*((t-1)*t)/2;
    else
        min=m*((t-1)*t)/2+r*t;
    printf("%I64d %I64d
",min,max);
    return 0;
}