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:
import sys
def ints_input():
    return [int(i) for i in sys.stdin.readline().strip("\n").split(" ")]


def print_list(arr):
    sys.stdout.writelines(str(x)+" " for x in arr)
    sys.stdout.write("\n")


def fast_input(type=str):
    return type(sys.stdin.readline().strip("\n"))

memo = {}
def d(n):
    ans = 0
    if n in memo:
        return memo[n]
    r = 1
    while r*r <= n:
        if n % r ==0:
            if n == r*r:
                ans += 1
            else:
                
                ans += 2
        r +=1 
    memo[n] = ans
    return ans


res = 0
a, b, c = ints_input()
for i in range(1, a+1):
    for j in range(1, b+1):
        for k in range(1, c+1):
            res += d(i*j*k)

print(res)