Problem:
Score : 400 points

Problem Statement

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).

Sample Input 1

4

Sample Output 1

23

We have f(1)=1, f(2)=2, f(3)=2, and f(4)=3, so the answer is 1\times 1 + 2\times 2 + 3\times 2 + 4\times 3 =23.

Sample Input 2

100

Sample Output 2

26879

Sample Input 3

10000000

Sample Output 3

838627288460105

Watch out for overflows.

Solution in C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

long long cnt_divisors(long long num){
  long long cnt = 0;
  for(long long i = 1; i*i <= num ; i++){
    if(num%i==0){
      cnt += 2;
    }
    if(i*i==num){
      cnt --;
    }
  }
  return cnt;
}

int main(void){
  long long n;
  long long sum = 0;
  scanf("%lld",&n);
  for(long long k = 1; k <= n ; k++ ){
    sum += k*cnt_divisors(k);
  }
  printf("%lld",sum);
  return 0;
}