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 ≤ 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.util.HashMap;
import java.util.Scanner;

public class SimpleMathProb {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner input = new Scanner(System.in);
		
		int a = input.nextInt();
		int b = input.nextInt();
		int c = input.nextInt();
		
		HashMap<Long, Long> memo = new HashMap<Long, Long>();
		
		long sum = 0;
		for(long i = 1; i <= a; i++){
			for(long j = 1; j <= b; j++){
				for(long k = 1; k <= c; k++){
					if(memo.containsKey(i*j*k)){
						sum += memo.get(i*j*k);
						continue;
					}
					long val = countDivisors(i * j * k);
					memo.put(i*j*k, val);
					sum += val;
				}	
			}
		}
		System.out.println(sum);
	}
	
	static int countDivisors(long n) { 
		int count = 0;
        // Note that this loop runs till square root 
        for (int i=1; i<=Math.sqrt(n); i++) 
        { 
            if (n%i==0) 
            { 
                // If divisors are equal, print only one 
                if (n/i == i) 
                    count++; 
       
                else // Otherwise print both 
                    count += 2; 
            } 
        } 
        return count;
    } 
}
