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.
#include <bits/stdc++.h>
using namespace std;
static const int MOD = 1 << 30;
int a, b, c;
long long num_of_divisors(long long x) {
  long long res = 1;
  int coef = 0;
  while (x % 2 == 0) {
    coef++;
    x /= 2;
  }
  res = coef + 1;
  int fac = 3;
  while (fac * fac <= x) {
    coef = 0;
    while (x % fac == 0) {
      x /= fac;
      coef++;
    }
    fac += 2;
    res *= (coef + 1) % MOD;
  }
  if (x != 1) {
    res *= 2 % MOD;
  }
  return res;
}
int main() {
  cin >> a >> b >> c;
  long long res = 0;
  for (int i = 1; i <= a; ++i) {
    for (int j = 1; j <= b; ++j) {
      for (int k = 1; k <= c; ++k) {
        res = res % MOD + num_of_divisors(i * j * k) % MOD;
      }
    }
  }
  cout << res;
  return 0;
}
