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 math import sqrt, ceil
a, b, c = list(map(int, input().split()))
N = (a * b * c) + 1
def generate_prime(N, pp, prime):
    pp[1] = False
    for i in range(2, N):
        if pp[i]:
            prime.append(i)
            x = i ** 2
            while x < N:
                pp[x] = False
                x = x + i

prime = []
pp = [True] * N

generate_prime(N, pp, prime)

def divisor_count(num):
    ans = 1
    if num == 1:
        return 1
    
    for p in prime:
        factor = 0
        while num:
            if num % p == 0:
                factor += 1
                num = num // p
            else:
                ans = ans * (factor + 1)
                break
        if pp[num]:
            ans = ans * 2
            break
        if num == 1:
            break
    return ans
        
            
d = {}
res = 0
for i in range(1, a+1):
    for j in range(1, b+1):
        for k in range(1, c+1):
            t = i * j * k
            if t not in d:
                d[t] = divisor_count(t)
            res = res + d[t]
print(res)