Problem: 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

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;

public class Main implements Runnable
{
//    static int dx[]={-1,0,1,0};
//    static int dy[]={0,-1,0,1};
    static long mod=1000000007;
    
    public static void main(String[] args) {        
        new Thread(null, new Main(), "", 200 * 1024 * 1024).start();        
    }

    public void run() {
        Reader sc = new Reader(System.in);
        PrintWriter out=new PrintWriter(System.out);

        int n=sc.nextInt();
        int lp[]=lp(10000001);
        long r=1;
        for (int i = 2; i <= n; i++) {
        	r+=ds(i,lp)*i;
//        	db(i,r);
		}
        
        out.println(r);
     	out.flush();
    }
    
    static long ds(int x, int lp[]) {
    	int d=0,k;
    	long r=1;
    	while(x!=1) {
    		d=lp[x];k=0;
    		while(d==lp[x]) {
    			x/=d;
    			k++;
    		}
    		r*=k+1;
    	}
    	return r;
    }
    
    static int[] lp(int n) {
    	int lp[]=new int[n+1];
    	for(int i=2; i<=n; i+=2)lp[i]=2;
    	for(int i=3; i*i<=n; i+=2) {
    	    if(lp[i]==0){
    	        lp[i]=i;
    	        for (int t=i*i; t<=n; t+=i) {
    	            if(lp[t]==0)lp[t]=i;
    	        }
    	    }
    	}
    	for(int i=1; i<=n; i++)if(lp[i]==0)lp[i]=i;
    	return lp;
    }
        
    static void ret(String ans) {
    	System.out.println(ans);
    	System.exit(0);
    }

    static void ret(long ans) {
    	System.out.println(ans);
    	System.exit(0);
    }
    
    static boolean validpos(int x,int y,int r, int c){
        return x<r && 0<=x && y<c && 0<=y;
    }

    static void db(Object... os){
        System.err.println(Arrays.deepToString(os));
    }
}

class P {
	int x,y,c;
	P(int  x, int y, int c) {
	  this.x=x;
	  this.y=y;
	  this.c=c;
	}
}

//
//class P implements Comparable<P>{
//	int x,c;long s;
//	P(int  x, int c, long s) {
//	  this.x=x;
//	  this.c=c;
//	  this.s=s;
//	}
//
//    public int compareTo(P p){
//        return Long.signum(s-p.s);
//    }
//}

class Reader
{ 
    private BufferedReader x;
    private StringTokenizer st;
    
    public Reader(InputStream in)
    {
        x = new BufferedReader(new InputStreamReader(in));
        st = null;
    }
    public String nextString()
    {
        while( st==null || !st.hasMoreTokens() )
			try {
				st = new StringTokenizer(x.readLine());
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
        return st.nextToken();
    }
    public int nextInt()
    {
        return Integer.parseInt(nextString());
    }
    public int[] nextIntArray(int size) {
        int r[] = new int[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextInt(); 
        }
        return r;
    }
    public long[] nextLongArray(int size) {
        long r[] = new long[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextLong(); 
        }
        return r;
    }
    public char[] getCharSet() {
        return this.nextString().toCharArray();
    }    
    public long nextLong()
    {
        return Long.parseLong(nextString());
    }
    public double nextDouble() throws IOException
    {
        return Double.parseDouble(nextString());
    }
}