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 ≤ 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.
#include <bits/stdc++.h>
using namespace std;
bool ok[1000000 + 1];
int primes[1000000 + 1];
int numprimes = 2;
void sieve(void) {
  primes[0] = 2;
  primes[1] = 3;
  for (int z = 0; z <= 1000000; z++) ok[z] = false;
  for (int x = 1; x <= 1000; x++) {
    for (int y = 1; y <= 1000; y++) {
      long int n = (4 * x * x) + (y * y);
      if (n <= 1000000 && (n % 12 == 1 || n % 12 == 5)) ok[n] ^= true;
      n = (3 * x * x) + (y * y);
      if (n <= 1000000 && n % 12 == 7) ok[n] ^= true;
      n = (3 * x * x) - (y * y);
      if (x > y && n <= 1000000 && n % 12 == 11) ok[n] ^= true;
    }
  }
  for (int r = 5; r <= 1000; r++)
    if (ok[r])
      for (int i = r * r; i < 1000000; i += r * r) ok[i] = false;
  for (int a = 5; a < 1000000; a++) {
    if (ok[a]) {
      primes[numprimes++] = a;
    }
  }
}
int n_divisors(int n) {
  int divisors = 1;
  int i = 0, power;
  {
    for (; primes[i] * primes[i] <= n; ++i) {
      power = 0;
      while (!(n % primes[i])) {
        n /= primes[i];
        power++;
      }
      divisors *= power + 1;
    }
  }
  if (n > 1) {
    divisors *= 2;
  }
  return divisors;
}
int main() {
  sieve();
  int count = 0;
  int a, b, c;
  scanf("%d%d%d", &a, &b, &c);
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++)
      for (int k = 1; k <= c; k++) {
        count += n_divisors(i * j * k);
        if (count > 1073741824) count %= 1073741824;
      }
  cout << count << endl;
  return 0;
}
