Problem: 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:

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Find the sum modulo 1073741824 (230).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 2000).

Output

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

Examples

Input

2 2 2


Output

20


Input

4 4 4


Output

328


Input

10 10 10


Output

11536

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.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class primes {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] input = br.readLine().split(" ");
		int a = Integer.parseInt(input[0]);
		int b = Integer.parseInt(input[1]);
		int c = Integer.parseInt(input[2]);
		int[] divisores = new int[10000001];
		long total = 0;
		for(int i =1;i<=1000000;i++) for(int j =i;j<=1000000;j+=i) divisores[j]++;
		for(int i =1;i<=a;i++)for(int j =1;j<=b;j++)for(int k =1;k<=c;k++){
			total+=divisores[i*j*k];
		}
		System.out.println(total);
	}
}