For a positive integer X, let f(X) be the number of positive divisors of X.

Given a positive integer N, find \sum_{K=1}^N K\times f(K).

Constraints

* 1 \leq N \leq 10^7

Input

Input is given from Standard Input in the following format:


N


Output

Print the value \sum_{K=1}^N K\times f(K).

Examples

Input

4


Output

23


Input

100


Output

26879


Input

10000000


Output

838627288460105
import java.util.*;
import java.util.Scanner;
import java.io.*;
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer;

import javax.lang.model.util.ElementScanner6; 

public class Main
{
    static class FastReader
    { 
        BufferedReader br; 
        StringTokenizer st; 

        public FastReader() 
        { 
            br = new BufferedReader(new
                    InputStreamReader(System.in)); 
        } 

        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 

        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 

        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 

        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 

        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
    
    

    public static void main(String args[])
    {  
        FastReader in=new FastReader();
        int t=1;
        //t=in.nextInt();
        while(t-->0)
        {
            long n=in.nextInt();
            long ans=0;
            for(long i=1;i<=n;i++)
            {
                long y=(long)Math.floor(n/(double)i);
                ans+=((y*(y+1))*i)/2;
            }
            System.out.println(ans);
        
        }

    }
}