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;

/**
 * @author artyom
 */
public class EasyNumberChallenge {
    private static final int MOD = 1073741824;

    private static final int MAX_PRIME_NUMBER = 100 * 100 * 100;
    private static final short[] primes = new short[MAX_PRIME_NUMBER + 1];

    static {
        generatePrimes();
    }

    private static void generatePrimes() {
        for (int i = 2, n = MAX_PRIME_NUMBER / 2; i <= n; i++) {
            for (int j = 2 * i; j <= MAX_PRIME_NUMBER; j += i) {
                primes[j]++;
            }
        }
    }

    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        long a = sc.nextLong(), b = sc.nextLong(), c = sc.nextLong();
        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 += d(i * j * k) % MOD;
                }
            }
        }
        System.out.println(sum);
    }

    private static long d(int num) {
        return num == 1 ? 1 : 2 + primes[num];
    }
}