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 functools import lru_cache
from math import sqrt,ceil
@lru_cache(None)
def countDevisor(n):
	count=0
	for i in range(int(sqrt(n))):
		if n%(i+1)==0:
			if n/(i+1)==(i+1):
				count+=1
			else:
				count+=2
	return(count)


a,b,c=list(map(int,input().split()))
modulo=2**30
ans=0
for i in range(1,a+1):
	for j in range(1,b+1):
		for k in range(1,c+1):
			ans+=countDevisor(i*j*k)
			#print(i*j*k,countDevisor(i*j*k))
print(ans%modulo)