You are given a positive integer n greater or equal to 2. For every pair of integers a and b (2 ≤ |a|, |b| ≤ n), you can transform a into b if and only if there exists an integer x such that 1 < |x| and (a ⋅ x = b or b ⋅ x = a), where |x| denotes the absolute value of x.

After such a transformation, your score increases by |x| points and you are not allowed to transform a into b nor b into a anymore.

Initially, you have a score of 0. You can start at any integer and transform it as many times as you like. What is the maximum score you can achieve?

Input

A single line contains a single integer n (2 ≤ n ≤ 100 000) — the given integer described above.

Output

Print an only integer — the maximum score that can be achieved with the transformations. If it is not possible to perform even a single transformation for all possible starting integers, print 0.

Examples

Input

4


Output

8

Input

6


Output

28

Input

2


Output

0

Note

In the first example, the transformations are 2 → 4 → (-2) → (-4) → 2.

In the third example, it is impossible to perform even a single transformation.

import java.io.*;
import java.util.*;

/**
 *  
 * @author akash_2ehuikg
 * Coding is love <3!
 */

public class D {
        static long sum(long i,long n){
            long ans=((n*(n+1))/2)-1;
            return ans;
        }
        
    public static void main(String args[]) {
    
        try {
            long n=nextLong();
            long ans=0;
            for (long i = 2; i <= n; i++) {
                if(n/i > 1){
                    ans+=(4*sum(i,n/i));
                 //   System.out.println(""+ans);
                }
            }
            
            bw.write(ans+"\n");
            bw.flush();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }


    //TEMPLATE
    
    //******************************************************************************
    // Fast I/O 
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
    static StringTokenizer st = null;
    static int mod=1000000007;

    static String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    
    static int nextInt() {
        return Integer.parseInt(next());
    }
    
    static long nextLong() {
        return Long.parseLong(next());
    }

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

    static int b_search(int l,int r,int A[],int key){
        int ans=0;
        l=0;r=1000000000;

        while(l<r){
            //System.out.println(ans);
            ans=(l+r)/2;
            long temp=0;
            for (int i = 0; i < A.length; i++) {
                temp+=(long)Math.ceil((double)A[i]/(double)ans);
            }

            if(temp<=key){
                r=ans;
            }
            else{
                l=ans+1;
                ans=r;
            }
        }

        return ans;
    }

//******************************************************************************
//Modulo and FastExpo
    static long modInverse(long a)
    {
       return power(a, mod - 2,mod);
    }
     
    // To compute x^y under modulo m
    static long power(long x, long y, int p)
    {
        // Initialize result
        long res = 1;     
        // Update x if it is more  
        // than or equal to p
        x = x % p; 
     while (y > 0)
        {
            // If y is odd, multiply x
            // with result
            if((y & 1)==1)
                res = (res * x) % p;
            // y must be even now
            // y = y / 2
            y = y >> 1; 
            x = (x * x) % p; 
        }
        return res;
    }
//******************************************************************************
//GCD
    static long gcd(long a, long b)
    {
        if (a == 0)
            return b;
         
        return gcd(b%a, a);
    }

    //******************************************************************************
    // Useful user datatype
    
    static class Pair{
        public long x;
        public long y;

        public Pair(long first, long second){
            this.x = first;
            this.y = second;
            }
        @Override
        public String toString() {
        return "(" + x + "," + y + ")";
        }
    }

    static Comparator csort=new Comparator<Pair>(){
                    public int compare(Pair a, Pair b) {
                        if(a.x < b.x)
                        return -1;
                        else if (a.x > b.x)
                        return 1;
                        else if(a.y < b.y)
                        return -1;
                        else if (a.y > b.y)
                        return 1;
                        else
                        return 0;
                    }
                };

    //******************************************************************************
    //N choose K
    public static long choose(long total, long choose){
        if(total < choose)
            return 0;
        if(choose == 0 || choose == total)
            return 1;
        return choose(total-1,choose-1)+choose(total-1,choose);
    }

    //******************************************************************************
    //Permutations
    // simply prints all permutation - to see how it works
    private static void printPermutations( Comparable[] c ) {
            System.out.println( Arrays.toString( c ) );
            while ( ( c = nextPermutation( c ) ) != null ) {
                    System.out.println( Arrays.toString( c ) );
            }
    }

    // modifies c to next permutation or returns null if such permutation does not exist
    private static Comparable[] nextPermutation( final Comparable[] c ) {
            // 1. finds the largest k, that c[k] < c[k+1]
            int first = getFirst( c );
            if ( first == -1 ) return null; // no greater permutation
            // 2. find last index toSwap, that c[k] < c[toSwap]
            int toSwap = c.length - 1;
            while ( c[ first ].compareTo( c[ toSwap ] ) >= 0 )
                    --toSwap;
            // 3. swap elements with indexes first and last
            swap( c, first++, toSwap );
            // 4. reverse sequence from k+1 to n (inclusive) 
            toSwap = c.length - 1;
            while ( first < toSwap )
                    swap( c, first++, toSwap-- );
            return c;
    }

    // finds the largest k, that c[k] < c[k+1]
    // if no such k exists (there is not greater permutation), return -1
    private static int getFirst( final Comparable[] c ) {
            for ( int i = c.length - 2; i >= 0; --i )
                    if ( c[ i ].compareTo( c[ i + 1 ] ) < 0 )
                            return i;
            return -1;
    }

    // swaps two elements (with indexes i and j) in array 
    private static void swap( final Comparable[] c, final int i, final int j ) {
            final Comparable tmp = c[ i ];
            c[ i ] = c[ j ];
            c[ j ] = tmp;
    }
    //******************************************************************************
    // DSU
    //Maintain an Array A with N elements and Array size for size of each set, intialize A[i]=i and size[i]=1

    static int root(int A[],int i){
        while(A[i]!=i){
            A[i]=A[A[i]];
            i=A[i];
        }
        return i;
    }

    static boolean find(int A[],int a,int b){
        if(root(A,a)==root(A,b))return true;
        else return false;
    }

    static void union(int A[],int size[],int a,int b){
        int ra=root(A,a);
        int rb=root(A,b);

        if(ra==rb)return;

        if(size[ra]<size[rb]){
            A[ra]=A[rb];
            size[rb]+=size[ra];
        }else{
            A[rb]=A[ra];
            size[ra]+=size[rb];
        }
    }

    //**************************************************************************
    //BinarySearch
    /* binary search for 5:
                v-- lower bound
         1 2 3 4 5 5 5 6 7 9
                      ^-- upper bound
        
            binary search for 8
                        v-- lower bound
       1 2 3 4 5 5 5 6 7 9
                        ^-- upper bound
    */
    //For finding greater than 
    static int upper_bound(int A[],int key){
        int first = 0;
        int last = A.length;
        int mid;

        while (first < last) {
            mid =first+(last - first)/2; 

            if (A[mid] <= key)  
                first = mid + 1; 
            else 
                last = mid;
        }

        return first;
    }
    
    //For finding greater than equal to
    static int lower_bound(int A[],int key){
        int first = 0;
        int last = A.length;
        int mid=0;
        
        while (first < last) {
            mid = first + ((last - first) >> 1); 
            if (A[mid] < key)  
                first = mid + 1; 
            else 
                last = mid;
        }
        
        return first;
    }
    //**************************************************************************
}
