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 sys
import math
input = sys.stdin.readline
from functools import cmp_to_key;

def pi():
    return(int(input()))
def pl():
    return(int(input(), 16))
def ti():
    return(list(map(int,input().split())))
def ts():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
f = [];
def factMod(n,m):
    global f;
    f = [1 for i in range(n+1)];
    f[0] = 1;
    for i in range(1,n+1):
        f[i] = (f[i-1]*i)%m;

def fact(n):
    global f;
    f = [1 for i in range(n)];
    for i in range(1,n):
        f[i] = f[i-1]*i;

def fast_mod_exp(a,b,m):
    res = 1;
    while b > 0:
        if b & 1:
            res = (res*a)%m;
        a = (a*a)%m;
        b = b >> 1;
    return res;

def inverseMod(n,m):
    return fast_mod_exp(n,m-2,m);

def ncrMod(n,r,m):
    if r == 0: return 1;
    if n < r: return 0;
    return ((f[n]*inverseMod(f[n-r],m))%m*inverseMod(f[r],m))%m;

def ncr(n,r):
    if r == 0: return 1;
    if n < r: return 0;
    return (f[n])/(f[r]*f[n-r]);
def nc2(n):
    return (n*(n-1))/2;

def main():
    D();

def D():
    n = pi();
    ans = 0;
    for i in range(2,n):
        ans += 4*i*(math.floor(n/i)-1);
    print(ans);
    
main();