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.*;

public class easynum {
    static int a, b, c;
    static long numDivisors[], sum;
    static final int mod = 1073741824;
	
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        //BufferedReader in = new BufferedReader(new FileReader("easynum.in"));
        
        StringTokenizer abc = new StringTokenizer(in.readLine());
        a = Integer.parseInt(abc.nextToken());
        b = Integer.parseInt(abc.nextToken());
        c = Integer.parseInt(abc.nextToken());
        numDivisors = new long[a*b*c+1];
        
        for (int i = 1; i <= a*b*c; i++) {
        	for (int j = 1; i*j <= a*b*c; j++) {
        		numDivisors[i*j] = (numDivisors[i*j]+1)%mod;
        	}
        }
        
        for (int i = 1; i <= a; i++) {
        	for (int j = 1; j <= b; j++) {
        		for (int k = 1; k <= c; k++) {
        			sum = (sum + numDivisors[i*j*k])%mod;
        		}
        	}
        }
        System.out.println(sum);
    }
}