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.Scanner;

public class Prob236B {
	public static void main(String[] Args) {
		int[] primes = new int[] { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,
				41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 };
		int[][] divisors = new int[101][primes.length];
		for (int i = 0; i < primes.length; i++) {
			int temp = primes[i];
			for (int power = 1; temp <= 100; temp *= primes[i], power++)
				for (int j = 1; j * temp <= 100; j++)
					divisors[j * temp][i] = power;
		}

		Scanner scan = new Scanner(System.in);
		int a = scan.nextInt();
		int b = scan.nextInt();
		int c = scan.nextInt();

		long temp = 0;
		long mod = 1073741824;
		for (int i = 1; i <= a; i++)
			for (int j = 1; j <= b; j++)
				for (int k = 1; k <= c; k++) {
					long d = 1;
					for (int p = 0; p < primes.length; p++)
						d = d
								* (divisors[i][p] + divisors[j][p]
										+ divisors[k][p] + 1) % mod;
					temp = (temp + d) % mod;
				}

		System.out.println(temp);
	}
}
