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 NumberChallenge {
    static int[] cache;

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);

        final int MOD = (1 << 30);

        int a = scan.nextInt();
        int b = scan.nextInt();
        int c = scan.nextInt();

        cache = new int[a*b*c + 1];

        long sum = 0;
        for (int i = 1; i <= a; i++) {
            for (int j = 1; j <= b; j++) {
                for (int k = 1; k <= c; k++) {
                    sum = (sum + d(i * j * k)) % MOD;
                }
            }
        }

        System.out.println(sum);
    }

    public static int d(int n) {
        if (cache[n] != 0)
            return cache[n];

        int num = n;
        int count = 1;
        while (num % 2 == 0) {
            num /= 2;
            count++;
        }

        for (int i = 3; i <= num; i += 2) {
            int times = 0;
            while (num % i == 0) {
                num /= i;
                times++;
            }
            count *= (times + 1);
        }
        cache[n] = count;
        return cache[n];
    }
}