CodeForces 236B Easy Number Challenge

Description

Let's denote d(n) as the number of divisors of a positive integer n. You are given three integers a, b and c. Your task is to calculate the following sum:

 

 

Find the sum modulo 1073741824(2³⁰).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 100).

Output

Print a single integer — the required sum modulo 1073741824(2³⁰).

Sample Input

Input

2 2 2

Output

20

Input

5 6 7

Output

1520


//预处理出1~1000000数的因子个数。

#include <iostream>
#include <cstdio>
using namespace std;
int mod = 1073741824;
long x[1000001];
int main()
{
	for (int i=1; i<1000001; i++)
	for (int j=i; j<1000001; j+=i)
		x[j]++;
	int a,b,c;
	long sum=0;
	while(cin>>a>>b>>c)
	{
	    for (int i=1; i<=a; i++)
	    {
          for (int j=1; j<=b; j++)
           {
               for (int k=1; k<=c; k++)
               {
                   sum += x[i*j*k];

               }
           }
	    }
	    if(sum > mod)
            {
              sum %= mod;
            }
		printf("%ld",sum);
		cout<<endl;
	}


	return 0;
}