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.Hashtable;
import java.util.Scanner;

public class EasyNumberChallenge {
    static Hashtable<Integer, Integer> divisorsCount = new Hashtable<>();

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int a = s.nextInt();
        int b = s.nextInt();
        int c = s.nextInt();
        System.out.println(getRequiredSumModulo(a, b, c));
    }

    private static long getRequiredSumModulo(int firstNumber, int secondNumber, int thirdNumber) {
        long total = 0;
        for (int i = 1; i <= firstNumber; i++) {
            for (int j = 1; j <= secondNumber; j++) {
                for (int k = 1; k <= thirdNumber; k++) {
                    total += getDivisorsCount(i * j * k);
                }
            }
        }
        return total % 1073741824;
    }

    private static long getDivisorsCount(int product) {
        if (divisorsCount.containsKey(product)) {
            return divisorsCount.get(product);
        }
        divisorsCount.put(product, (int) getDivisors(product));
        return divisorsCount.get(product);
    }

    private static long getDivisors(int number) {
        int divisorsCount = 0;
        for (int i = 1; i <= Math.sqrt(number); i++) {
            if (number % i == 0)
                divisorsCount = number / i == i ? divisorsCount + 1 : divisorsCount + 2;
        }
        return divisorsCount;
    }
}