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:
pr=[1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
cd = {}
def d(n):
    if n in cd:
        return cd[n]
    t,i=1,2
    k=n

    while i * i <= k:
        if k % i == 0:
            k /= i
            c = 2
            while k % i == 0:
                k /= i
                c += 1
            t *= c
        i += 1
    if k > 1:
        t *= 2
    cd[n] = t
    return t
i,j,m=map(int, raw_input().split())
r=0
for x in xrange(1,i+1):
    for y in xrange(1, j+1):
        for z in xrange(1, m+1):
            #l=len(set(a*s*w for a in d(x) for s in d(y) for w in d(z)))
            #
            #print len(d(x)|d(y)|d(z))
            r += d(x*y*z)
print r%1073741824