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.*;

public class Main {
    public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();

		// 計算
		long result = 0;
		int[] list = new int[n+1];
		Arrays.fill(list, 1);
		for(int i = 2; i <= n; i++){
    	    for(int j = i; j <= n; j += i){
    	        list[j]++; 
			}
		}
		
		for(int i = 1; i <= n; i++){
    	    result += (long)i*list[i];
		}
		
		// 出力
		System.out.println(result);
	}

	public static class Scanner {
		private BufferedReader br; private StringTokenizer tok;
		public Scanner(InputStream is) throws IOException { br = new BufferedReader(new InputStreamReader(is));}
		private void getLine() throws IOException { while(!hasNext()) tok = new StringTokenizer(br.readLine());}
		private boolean hasNext() { return tok != null && tok.hasMoreTokens();}
		public String next() throws IOException { getLine(); return tok.nextToken();}
		public int nextInt() throws IOException { return Integer.parseInt(next());}
		public long nextLong() throws IOException { return Long.parseLong(next());}
		public double nextDouble() throws IOException { return Double.parseDouble(next());}
	}
}