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 math
def getdivisors(n):
    #divisors = e1+1 *  e2+1 *.......ek+1
    divisors = 1
    temp = 0
    while n%2==0:
        n//=2
        temp+=1
    divisors*=temp+1
    temp = 0
    i=3
    while i<math.sqrt(n)+1:
        while n%i==0:
            temp+=1
            n//=i
            if n%i!=0:
                divisors*=temp+1
                temp =0
        i+=2
    if n>2:
        divisors*=2
    return divisors


a,b,c = [int(x) for x in input().split()]
hashmap = {}
summ=0
for i in range(1,a+1):
    for j in range(1,b+1):
        for k in range(1,c+1):
            x = i*j*k
            if x in hashmap:
                summ+=hashmap[x]
            else:
                hashmap[x]=getdivisors(x)
                summ+=hashmap[x]
print(summ)