import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    static long sumOfGroup(int k) {
        // Return the sum of the elements of the k'th group.
        /* which number of the sequence do i start at?
        if k == 1, then i get the first num
        k == 2, i get the 2 num
        k == 3, i get the 4th num
        k == 4, start at 7th
        (k-1) + (k-2) + 1, plus another one
        this is (k-1)(k)/2 + 1
        the actual number i start at is double this, then minus one
        or take out the plus one and add it after the doubling
        */
        /*
        long sum = 0;
        long currNum = (k-1) * k + 1;
        for (int i = 0; i < k; i++, currNum += 2) {
            sum += currNum;
        }
        return sum;
        
        ok so that was pretty good, but i can do better
        once i have the starting number, i can calculate the average number and then multiply by k
        if there are 2 numbers in the set, i get the average by adding 1
        if 3 numbers, average by adding 2
        add k-1 to get the average
        */
        return (long)k * k * k;
        //return (long)((k-1) * k + k) * (long)k;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int k = in.nextInt();
        long answer = sumOfGroup(k);
        System.out.println(answer);
        in.close();
    }
}