235_E. Number Challenge
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:

<image>

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.
```python
primes = []
d = {}
for i in range(2, 100):
    if i not in d:
        primes.append(i)
        j = 2
        while i * j < 100:
            d[i*j] = 1
            j += 1

def solve(x,y,z):
    d = {}
    for p in primes:
        d[p] = 0

    for p in primes:
        if p > x and p > y and p > z:
            break
        while x%p == 0:
            x /= p
            d[p] += 1
        while y%p == 0:
            y /= p
            d[p] += 1
        while z%p == 0:
            z /= p
            d[p] += 1
    ans = 1
    for p in primes:
        ans *= d[p] + 1
    return ans

a,b,c = map(int,input().split(' '))
ans = 0
seen = {}

for i in range(1, a+1):
    for j in range(1, b+1):
        for k in range(1, c+1):
            p = i*j*k
            if p in seen:
                ans = (ans + seen[p]) % 1073741824
            else:
                temp = solve(i,j,k)
                ans = (ans + temp) % 1073741824
                seen[p] = temp

print(ans)



```