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 dari library //
import java.util.*; // import dari library //
import java.text.*; // import dari library //
import java.math.*; // import dari library //
import java.util.regex.*; // import dari library //

public class RESPONSI_PSDA_478B {

    public static BigInteger n(long x) { // mendeklarasikan method BigInteger n dengan parameter x tipe data long //
        return BigInteger.valueOf((x) * (x - 1) / 2);  //mengembalikan nilai BigInteger.valueOf((x) * (x - 1) / 2)//
    }

    public static void main(String[] args) throws IOException { // Mendeklarasikan method main dengan tipe akses public dengan membangkitkan eksepsi //
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); // perintah untuk input data melalui objek br  //
        String line = br.readLine(); // membaca nilai dari br //
        String[] parts = line.split(" "); // memisahkan data yang diinputkan //
        String first = parts[0]; // data pertama akan memasuki indeks ke 0 //
        String second = parts[1]; // data kedua akan memasuki indeks ke 1 //
        int p = Integer.parseInt(first); // mendeklarasikan variabel p = diambil dari nilai data pertama //
        int t = Integer.parseInt(second); // mendeklarasikan variabel t = diambil dari nilai data kedua  //
        BigInteger bb = n(p - t + 1); // mendeklarasikan nilai   BigInteger bb = n(p - t + 1) //
        BigInteger ll; 
        if (p % t == 0) { // jika nilai modulo p dengan t ==0 //
            ll = n(p / t).multiply(BigInteger.valueOf(t)); // maka nilai ll adalah n(p / t).multiply(BigInteger.valueOf(t)) //
        } else if (p - t == 1) { // selain itu jika nilai p -t =1 //
            ll = BigInteger.valueOf(1); // maka nilai ll adalah   BigInteger.valueOf(1) //
        } else { // jika kondisinya tidak memenuhi semua //
            int up = p / t + 1; // mendeklarasikan variabel up =  p / t + 1;//
            int down = p / t;   // mendeklarasikan variabel down =  p / t 
            int n1 = 0;  // mendeklarasikan variabel n1=0 //
            while (p / t < up) { // selama nilai p /t < up //
                p -= down; // maka nilai p -=down //
                t -= 1; // nilai t-=l //
                n1 += 1; //nilai n1 += 1 //
            }
            ll = n(down).multiply(BigInteger.valueOf(n1)).add(n(up).multiply(BigInteger.valueOf(t)));
        } // nilai ll adalah  n(down).multiply(BigInteger.valueOf(n1)).add(n(up).multiply(BigInteger.valueOf(t)));//
        System.out.println(ll + " " + bb); // menampilkan nilai ll didapat dari argumen nilai bb //

    }
}
