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:
def inp():
    return map(int, stdin.readline().split())


def factorize(n):  # o(sqr(n))
    c, ans = 1, 0
    while (c * c < n):
        if n % c == 0:
            ans += 2
        c += 1

    if c * c == n:
        ans += 1
    return ans


from sys import *
from collections import *

a, b, c = inp()
mem, ans = defaultdict(int), 0

for i in range(1, a + 1):
    for j in range(1, b + 1):
        for k in range(1, c + 1):
            pro = i * j * k
            if mem[pro] == 0:
                mem[pro] = factorize(pro)
            ans += mem[pro]
print(ans)