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:

<image>

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

2 2 2


Output

20


Input

5 6 7


Output

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.
import java.io.*;

import java.util.Arrays;

public class EasyNumberChallenge
{ 
	static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); 
	
	static int numberOfDivisors(int n)
	{
		int count = 0;
		for (int i = 1; i <= Math.sqrt(n); i++)
		{
			if (n % i == 0)
			{
				count += 2;
				if (i * i == n)
					count--;
			}
		}
		return count;
	}

	public static void main (String [] args) throws IOException
	{
		int [] nums = Arrays.stream(reader.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();

		int [] divisors = new int[1000001];

		int answer = 0;
		for (int i = 1; i <= nums[0]; i++)
		{
			for (int j = 1; j <= nums[1]; j++)
			{
				for (int k = 1; k <= nums[2]; k++)
				{
					int product = (i*j*k);
					int toAdd = (divisors[product] != 0) ? divisors[product] : numberOfDivisors(i*j*k);
					divisors[product] = toAdd;
					answer += toAdd;
				}
			}
		}
		System.out.println(answer);
	}
}
