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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {

    private 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(final String[] args) throws IOException {
        FastReader in = new FastReader();
        PrintWriter out = new PrintWriter(System.out, true);

        int n = in.nextInt();
        int[] divisors = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= n; j += i) {
                divisors[j]++;
            }
        }

        long answer = 0;
        for (int i = 1; i <= n; i++) {
            answer += (long) i * divisors[i];
        }
        out.println(answer);
    }
}
