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:
abc=input().split(' ')
a=int(abc[0])
b=int(abc[1])
c=int(abc[2])

def product(f1,f2,f3):
	fout=[]
	for i in range(len(f1)):
		fout.append(f1[i]+f2[i]+f3[i])
	return fout
def no_of_factors(fin):
	n=1
	for inn in fin:
		n*=(inn+1)
	return n

primes=[2,3,5,7]
for i in range(11,101):
    bhul=True
    for j in primes:
        if(i%j==0):
            bhul=False
            break
    if(bhul):
        primes.append(i)

compendium=[]
for i in range(1,101):
	factors=[0]*len(primes)
	for pp in range(len(primes)):
		di=int(i)
		while di%primes[pp]==0:
			di/=primes[pp]
			factors[pp]+=1
	compendium.append(factors)

q=0
iterations=0
for i in range(a):
	for j in range(b):
		for k in range(c):
			q+=no_of_factors(product(compendium[i],compendium[j],compendium[k]))

print(q)