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 ≤ 100).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Examples

Input

2 2 2


Output

20


Input

5 6 7


Output

1520

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.
a, b, c = map(int, input().split())
primes = []
for i in range(2, 101):
    for j in range(2, i):
        if(i%j==0):
            break
    else:
        primes.append(i)

divisors = []
for i in range(1, 101):
    divisors.append([0]*len(primes))
    n = i
    for j in range(len(primes)):
        while(n>0 and n%primes[j]==0):
            n = n//primes[j]
            divisors[i-1][j]+=1

r = 0
for i in range(1, a+1):
    for j in range(1, b+1):
        for k in range(1, c+1):
            d = 1
            ijk_divs = [0]*len(primes)
            for zzz in range(len(primes)):
                d *= divisors[i-1][zzz]+divisors[j-1][zzz]+divisors[k-1][zzz]+1
            #print(i, j, k, d)
            r += d%1073741824

print(r)