B. Easy Number Challenge（有技巧的枚举）

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).

Examples

Input

Copy

2 2 2

Output

Copy

20

Input

Copy

5 6 7

Output

Copy

1520

Note

For the first example.

• d(1·1·1) = d(1) = 1;
• d(1·1·2) = d(2) = 2;
• d(1·2·1) = d(2) = 2;
• d(1·2·2) = d(4) = 3;
• d(2·1·1) = d(2) = 2;
• d(2·1·2) = d(4) = 3;
• d(2·2·1) = d(4) = 3;
• d(2·2·2) = d(8) = 4.

So the result is 1 + 2 + 2 + 3 + 2 + 3 + 3 + 4 = 20.

比赛时超时了好几次，最后开了个数组保存已经计算过的，就AC了。

AC Code：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

 没有想到还可以这样写，orz,巧妙呐

ACCode：

#include 
using namespace std;
int d[1000010];
int main()
{
	
	int a,b,c,i,j,k,s=0;
	
	for(i = 1;i<=1000000;i++)
    for(j = i;j<=1000000;j+=i)//i为因子，j为i的倍数
	 d[j]++;//预处理
	 
	cin>>a>>b>>c;
	
	for(i=1;i<=a;i++)
	for(j=1;j<=b;j++)
	for(k=1;k<=c;k++)
	{
		s+=d[i*j*k];
	}
	cout<