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 Maxn = 109;
const int mod = 1073741824;
vector<int> p;
bool is_prime(int n) {
  for (int i = 2; i < n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
void make_prime() {
  for (int i = 2; i < 1000; i++) {
    if (is_prime(i)) {
      p.push_back(i);
    }
  }
  return;
}
int dev(int n) {
  int i = 0, d = 1;
  while (n > 1) {
    int c = 1;
    while (n % p[i] == 0) {
      n /= p[i];
      c++;
    }
    d *= c;
    i++;
    if (i >= p.size()) {
      return d * 2;
    }
  }
  return d;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  make_prime();
  int a, b, c;
  long long s = 0;
  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++) {
        s = (s + dev(i * j * k)) % mod;
      }
    }
  }
  cout << s;
  return 0;
}
