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:
mod=(1<<30)  #dual shift means two raise to
d={}
def div(x):
    if x in d:
        return d[x]
    res, p, q = 1, 2, x
    while p * p <= x:
        r = 1
        while x % p == 0:
            x /= p
            r += 1
        res = (res * r) % mod
        p += 1
    if x > 1:
        res = (res * 2) % mod
    d[q] = res
    return res
a,b,c=sorted(map(int,input().split()))
s=0
for i in range(1,a+1):
    for j in range(1,b+1):
        for k in range(1,c+1):      #highly inefficient
            s+=div(i*j*k)
print(s)