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.util.Scanner;
public class EasyNumberChallange {
    static int div(int input)
    {
         if (input == 1) {
                return 1;
            }
            int maxD = (int) Math.sqrt(input);
            int sum = 0;
            for (int i = 1; i <= maxD; i++) {
                if (input % i == 0) {
                    
                    {
                        if(input/i==i)
                        {
                            sum++;
                        }
                        else
                        {
                            sum=sum+2;
                        }
                    }
                   
            }

            
        
             }
            return sum;
    }
    
    public static void main(String arg[])
    {
        Scanner sc=new Scanner(System.in);
        int a=sc.nextInt();
        int b=sc.nextInt();
        int c=sc.nextInt();
        int ar[]=new int[10000001];
        
            
        long s=0;
        for(int i=1;i<=a;i++)
        {
            for(int j=1;j<=b;j++)
            {
                for(int z=1;z<=c;z++)
                {
                    int x=i*j*z;
                    
                    if(ar[x]==0)
                    {
                        ar[x]=div(x);
                        s=(s+ar[x])%1073741824;
                    }
                    else
                    {
                        s=(s+ar[x])%1073741824;
                    }
                }
            }
        }
        System.out.println(s);
       
    }
}