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;
int primes[100000];
bool checkprime(int a) {
  int i = 1;
  while (primes[i] <= sqrt(a)) {
    if (a % primes[i] == 0) {
      return (false);
      goto qwe;
    } else
      i++;
  };
  return (true);
qwe:;
}
void primemaker() {
  primes[0] = 2;
  primes[1] = 3;
  int q = 5;
  int j = 2;
  while (j < 100000) {
    if (checkprime(q)) {
      primes[j] = q;
      j++;
      q = q + 2;
    } else
      q = q + 2;
  };
}
int div(int a) {
  int counter = 1, i = 0, divs = 0;
  while (a != 1) {
    while ((a % primes[i] == 0)) {
      a /= primes[i];
      divs++;
    }
    counter *= (divs + 1);
    i++;
    divs = 0;
  }
  return counter;
}
int main() {
  primemaker();
  int mod = 1073741824, ans = 0, a, b, c;
  cin >> a >> b >> c;
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        ans += div(i * j * k);
        ans %= mod;
      }
    }
  }
  cout << ans;
}
