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.Arrays;
import java.util.Scanner;


public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		new TaskB().solve(in.nextInt(), in.nextInt(), in.nextInt());
		in.close();
	}
}

class TaskB {

	boolean[] isPrime;
	int[] primes;
	int n = 0;
	int[] notedDivisiors;
	
	TaskB() {
		notedDivisiors = new int[1500000];
		int count = 0;
		isPrime = new boolean[1000];
		Arrays.fill(isPrime, true);
		isPrime[0] = isPrime[1] = false;
		for (int i = 2; i < isPrime.length; ++i) {
			if (isPrime[i]) {
				++count;
				for (int ii = 2 * i; ii < isPrime.length; ii += i)
					isPrime[ii] = false;
			}
		}
		primes = new int[count];
		count = 0;
		for (int i = 2; i < isPrime.length; ++i) 
			if(isPrime[i])
				primes[count++] = i;		
	}

	void solve(int a, int b, int c) {
		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++) {
					n = i*j*k;
					sum = (sum + this.d()) % mod;
				}
		System.out.print(sum);
	}
	
	int[] minDiv (int index) {
		int[] returned = new int[2];
		while (n % primes[index] != 0) {
			++index;
		}
		returned[0] = primes[index];
		returned[1] = index;
		return returned; 
	}
	
	int valueOf (int p) {
		int value = 0;
		while (n % p == 0) {
			n /= p;
			++value;
		}
		return value;
	}
	
	int d() {
		if (notedDivisiors[n] != 0)
			return notedDivisiors[n];
		int tempN = n;
		int result = 1;
		int[] temp = new int[2];
		temp[0] = temp[1] = 0;
		while (n > 1) {
			temp = this.minDiv(temp[1]);
			result *= 1 + this.valueOf(temp[0]);
		}
		notedDivisiors[tempN] = result;
		return result;
	}

}