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:

<image>

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:
import sys
input = lambda: sys.stdin.readline().strip("\r\n")
import math


def countDivisors(n):
    cnt = 0
    for i in range(1, (int(math.sqrt(n))) + 1):
        if n % i == 0:
            if int(n / i) == i:
                cnt = cnt + 1
            else:
                cnt = cnt + 2
    return cnt

d = dict()

a, b, c = map(int, input().split())
ans = 0
for i in range(1, a+1):
    for j in range(1, b+1):
        for k in range(1, c+1):
            if i * j * k not in d:
                d[i*j*k] = countDivisors(i*j*k)
                ans += d[i*j*k]
            else:
                ans += d[i*j*k]
print(ans)