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:
#!/usr/bin/env python3
import sys


primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
dp = {}

for i in range(len(primes)):
	dp[primes[i]] = 2

def factors(n):
	if n in dp:
		return dp[n]
	k = n
	s = 1
	for i in range(len(primes)):
		if (n%primes[i] == 0):
			c = 0
			while (n% primes[i] ==0) and n>=primes[i]:
				c += 1
				n = n//primes[i]
			s *= c+1
	dp[k] = s
	return s

if __name__ == "__main__":
	a, b, c = map(int, sys.stdin.readline().split())
	s = 0
	for i in range(1, a+1):
		for j in range(1, b+1):
			for k in range(1, c+1):
				s += factors(i*j*k)
	print(s%1073741824)