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 PYTHON3):

Solution:
from collections import defaultdict


def divisorGenerator(n):
    large_divisors = []
    for i in range(1, int((n ** 0.5) + 1)):
        if n % i == 0:
            yield i
            if i * i != n:
                large_divisors.append(n / i)
    for divisor in reversed(large_divisors):
        yield divisor


a, b, c = map(int, input().split())
number_of_divisors = defaultdict()
result = 0

for i in range(1, a + 1):
    for j in range(1, b + 1):
        for k in range(1, c + 1):
            x = i * j * k
            if x in number_of_divisors:
                d = number_of_divisors[x]
            else:
                d = len(list(divisorGenerator(x)))
                number_of_divisors[x] = d
            result = (result + d) % 1073741824

print(result)