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:
lis = list(map(int, input().split()))
sumall = 0
dic = {}

modulo = 1073741824
def number_of_divisors(a):
    if a in dic:
        return dic[a]
    res, i, t = 1, 2, a
    while i * i <= a:
        r = 1
        while a % i == 0:
            a /= i
            r += 1
        res = (res * r) % modulo
        i += 1
    if a > 1:
        res = (res * 2) % modulo
    dic[t] = res
    return res


for i in range(1, lis[0] + 1):
    for j in range(1, lis[1] + 1):
        for k in range(1, lis[2] + 1):
            s = (i, j, k)
            sumall += number_of_divisors(i * j * k)
print(sumall % modulo)