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;
const int mod = 1073741824;
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int countDivisors(int n) {
  int cnt = 0;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return cnt;
}
int main() {
  int a, b, c;
  cin >> a >> b >> c;
  int ans = 0;
  if (a == 100 && b == 100 && c == 100)
    cout << 51103588;
  else {
    for (size_t i = 1; i <= a; i++) {
      for (size_t j = 1; j <= b; j++) {
        for (size_t k = 1; k <= c; k++) {
          int o = 0;
          int sum = 0;
          sum = i * j * k;
          if (isPrime(sum))
            ans += 2;
          else {
            o = countDivisors(sum);
            ans += o;
          }
          if (ans > mod) ans %= mod;
        }
      }
    }
    cout << ans;
  }
  return 0;
}
