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 kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.

Examples

Input

5 1


Output

10 10


Input

3 2


Output

1 1


Input

6 3


Output

3 6

Note

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.
from math import *
from copy import *
from string import *				# alpha = ascii_lowercase
from random import *
from sys import stdin
from sys import maxsize
from operator import *				# d = sorted(d.items(), key=itemgetter(1))
from itertools import *
from collections import Counter		# d = dict(Counter(l))
import math

def bin1(l,r,k,t,b,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		v=k**mid
		if(v==val):
			return v
		elif(v>val):
			ans=mid
			return bin1(mid+1,r,k,t,b,val,ans)
		else:
			return bin1(l,mid-1,k,t,b,val,ans)
		
def bin2(l,r,k,t,b,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		v=t*(k**mid)+b*(mid)
		if(v==val):
			return v
		elif(v>val):
			ans=mid
			return bin2(l,mid-1,k,t,b,val,ans)
		else:
			return bin2(mid+1,r,k,t,b,val,ans)

def SieveOfEratosthenes(n): 
      
    # Create a boolean array "prime[0..n]" and initialize 
    #  all entries it as true. A value in prime[i] will 
    # finally be false if i is Not a prime, else true. 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
          
        # If prime[p] is not changed, then it is a prime 
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    return prime
def bin(l,r,ll,val):
	if(l>r):
		return -1
	else:
		mid=(l+r)//2
		if(val>=ll[mid][0] and val<=ll[mid][1]):
			return mid
		elif(val<ll[mid][0]):
			return bin(l,mid-1,ll,val)
		else:
			return bin(mid+1,r,ll,val)
def deci(n):
	s=""
	while(n!=0):
		if(n%2==0):
			n=n//2
			s="0"+s
		else:
			n=n//2
			s="1"+s
	return s
def diff(s1,s2):
	if(len(s1)<len(s2)):
		v=len(s1)
		while(v!=len(s2)):
			s1="0"+s1
			v=v+1
	else:
		v=len(s2)
		while(v!=len(s1)):
			s2="0"+s2
			v=v+1
	c=0
	for i in range(len(s1)):
		if(s1[i:i+1]!=s2[i:i+1]):
			c=c+1
	return c
from sys import stdin, stdout 
def fac(a,b):
	v=a
	while(a!=b):
		v*=a-1
		a=a-1
	return v
def bino(l,r,n):
	if(l>r):
		return -1
	else:
		mid=(l+r)//2
		val1=math.log((n/mid)+1,2)
		val2=int(val1)
		if(val1==val2):
			return val1
		elif(val1<1.0):
			return bino(l,mid-1,n)
		else:
			return bino(mid+1,r,n)

if __name__ == '__main__':
	n,m=map(int,input().split(" "))
	
	v1=n//m
	v2=m-(n%m)
	v11=v2*(v1*(v1-1))//2
	v12=(n%m)*(v1*(v1+1))//2
	# print(v11,v12)
	ans1=v11+v12
	ans2=n-(m-1)
	ans2=(ans2*(ans2-1))//2
	print(ans1,ans2)