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.
//package codeforces;

import java.util.Scanner;

/**
 * Created by nitin.s on 02/04/16.
 */
public class EasyNumberChallenge {
    public static int[] was=new int[1300000];
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int a = in.nextInt(), b = in.nextInt(), c = in.nextInt();
        final int mod = 1073741824;
        int 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);
    }

    private static int d(int n) {
        if(was[n] != 0) return was[n];
        int result = 0;
        int sqrtN = (int) Math.sqrt(n);
        for(int i = 1; i <= sqrtN; ++i) {
            if(n % i == 0)
                ++result;
        }
        if(sqrtN * sqrtN == n) {
            result = 2 * result - 1;
        } else {
            result = 2 * result;
        }
        was[n] = result;
        return result;
    }
}
