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 PYTHON):

Solution:
from sys import stdin ,stdout
from os import path
rd = lambda:stdin.readline().strip()
wr = stdout.write
if(path.exists('input.txt')):
    stdin = open("input.txt","r")
import time 
#------------------------------------
arr = list(map(int,rd().split()))
mylist = []
sum  = 0 
m = (arr[0]*arr[2]*arr[1])
dn = [1]*(m+1)
for i in range(2,m+1):
	for j in range(i, m+1, i ):
		dn[j] +=1
def d(n,counter):
	global sum,arr,dn
	if counter >= 3:
		sum+=dn[n]
	else: 
		for i in range(1,arr[counter]+1):
			d(i*n,counter+1)
d(1,0)
print(sum%1073741824)