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.*;
public class Ladder81 {

	public static void main(String[] args)throws IOException {
		BufferedReader ob=new BufferedReader(new InputStreamReader(System.in));
		String s[]=ob.readLine().split(" ");
		int a=Integer.parseInt(s[0]);
		int b=Integer.parseInt(s[1]);
		int c=Integer.parseInt(s[2]);
		long sum=0;
		long mod=(long)Math.pow(2,30);
		long div[]=new long[a*b*c+1];
		for(int i=1;i<=a;i++) {
			for(int j=1;j<=b;j++) {
				for(int k=1;k<=c;k++) {
					//System.out.print("i,j,k= "+i+","+j+","+k+" ");
					int x=i*j*k;
					if(div[x]==0)
						div[x]=divisor(x);
					
					sum=sum+div[x];
					sum=sum%mod;
					//System.out.print("sum= "+sum+" ijk= "+i*j*k+" ");
				}
				//System.out.println();
			}
		}
		System.out.println(sum);
	}
	static long divisor(int n) 
    { 	long a=0;
        for (int i=1;i<=Math.sqrt(n);i++) {
        	if(n%i==0) { 
                if (n/i==i) 
                	a++;
                else
                	a=a+2;
             } 
        }
        return a;
    } 
}