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.io.BufferedReader; 
import java.io.*;
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer; 
import java.util.*;
import java.math.*;
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[])
    {
        Scanner sc=new Scanner(System.in);
        PrintWriter pw=new PrintWriter(System.out);
        int t=1;//sc.nextInt();
        long ans = 0;
        long  a[]=new long[10000001];
        while(t-->0)
        {
        int n=sc.nextInt();
        pop(a,n);
        for(int i=1;i<=n;i++)
        {
            ans+=i*a[i];
        }
        pw.println(ans);
        }
           pw.close();
    }
    static void pop(long a[],int n)
    {
        for (int i = 1; i <= n; i++) {
            for (int j = i; j <= n; j += i) {
                a[j]++;
            }
        }
    }
}
  
