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 ≤ 100).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Examples

Input

2 2 2


Output

20


Input

5 6 7


Output

1520

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.
from sys import stdin, stdout
def insr():
    s = input()
    return(list(s[:]))


def inlt():
    return(list(map(int,input().split())))

inp = inlt()

sol = [0] * 1000001
def count_divisors(a):
    if sol[a] != 0:
        return sol[a]

    i, c = 1, 0
    while ((i * i) <= a):
        if a % i == 0:
            if (i * i) != a:
                c += 1

            c += 1

        i += 1

    sol[a] = c
    return c


s = 0
for i in range(1, inp[0] + 1):
    for j in range(1, inp[1] + 1):
        for k in range(1, inp[2] + 1):
            s += count_divisors(i * j * k)


print(s % 1073741824)



