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.
MAX = 101
MOD = 1073741824

nums = [True] * MAX
nums[0] = nums[1] = False
for i in xrange(2, 11):
    if not nums[i]:
        continue
    else:
        for j in xrange(i + i, MAX, i):
            nums[j] = False
primes = [i for i, e in enumerate(nums) if e]
plen = len(primes)
table = [[0] * plen for i in xrange(MAX)]
for i in xrange(2, MAX):
    x = i
    for j in xrange(plen):
        while True:
            d, m = divmod(x, primes[j])
            if m == 0:
                table[i][j] += 1
                x /= primes[j]
                if x == 1:
                    break
            else:
                break

a, b, c = map(int, raw_input().split())

memo = [0] * (a * b * c + 1)

s = 0
for i in xrange(1, a + 1):
    for j in xrange(1, b + 1):
        for k in xrange(1, c + 1):
            if memo[i * j * k] != 0:
                ts = memo[i * j * k]
            else:
                ts = 1
                for p in xrange(plen):
                    ts *= table[i][p] + table[j][p] + table[k][p] + 1
                    ts %= MOD
                memo[i * j * k] = ts
            s = (s + ts) % MOD
print s
