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.
import java.io.*;
import java.util.*;

import javax.swing.text.html.MinimalHTMLWriter;

import static java.lang.Math.abs;
import static java.lang.Math.max;
import static java.lang.Math.min;


public class Main {


    public static void main(String [] args){
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Integer readFromFile=new Integer(1);
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }
    static class TaskD {
        void madAdd(Map <Integer, Integer> map, int value){
            if(map.containsKey(value)){
                map.put(value, map.get(value)+1);
            }else{
                map.put(value, 1);
            }
        }
        //I have to try at least two ways to solve this problem, I'll make one of them with ;
        class Pair implements Comparable<Pair>{
            int fist;
            int second;
            Pair(int fist, int second){
                this.fist = fist;
                this.second = second;
            }
            @Override
            public int compareTo(Pair arg0) {
                return arg0.fist-fist;
            }
        }
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            long n = in.nextInt();
            long m = in.nextInt();
            long maxCouple = (long)((n-m+1)*((n-m+1)-1)/2);
            long minCouple = 0;
            if(n%m==0){
                minCouple=(long)(n/m*(n/m-1)/2)*m;
            }else{
                long mod = n%m;
                minCouple = (long)(n/m*(n/m-1)/2)*(m-mod);
                long value = n/m+1;///!!!!!!!
                minCouple+=(long)(value*(value-1)/2)*mod;
            }
            out.println(minCouple+" "+maxCouple);

        }
    }

    static class InputReader {
        BufferedReader br;
        StringTokenizer st;
        String st1;
        File file = new File("text.txt");
        public InputReader(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
            st = null;
        }
        public InputReader(int i) {
            try {
                br  = new BufferedReader(
                        new InputStreamReader(new FileInputStream(file)));
            } catch (FileNotFoundException e1) {
                System.out.println("File is not find");
            }
            st = null;
        }
        public String next(){
            while (st==null || !st.hasMoreTokens()){
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }
        public String nextLine(){
            try {
                st1 = new String(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return st1;
        }
        public int nextInt()  {return Integer.parseInt(next());}
        public long nextLong(){
            return Long.parseLong(next());
        }
        public Double nextDouble(){
            return Double.parseDouble(next());
        }
        public Byte nextByte() {
            return Byte.parseByte(next());
        }
        private int idx;
        public Character nextChar() {
            if(st1==null) {
                st1 = next();
                idx = 0;
            }
            if(idx!=(st1.length())-1){
                return st1.charAt(idx++);
            }else{
                char  c= st1.charAt(idx);
                st1 = null;
                return c;
            }
        }
        public void newFile() {
            try {
                FileWriter write = new FileWriter(file);
                write.write("something for cheaking");
                write.close();

            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        public void writeFile1(){
            try{
                FileWriter write = new FileWriter(file);
                write.append("100000"+" ");
                for(int i=0; i<100000; i++){
                    write.append(new Integer((int)(Math.random()*5000)).toString()+" ");
                }
                write.close();

            }catch(IOException e){
                e.printStackTrace();
            }
        }
    }
}