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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long mod = 1073741824;
  int x, y, z;
  cin >> x >> y >> z;
  map<int, int> m;
  map<int, int> repeat;
  int cntr = 0;
  for (int i = 1; i <= x; i++) {
    for (int j = 1; j <= y; j++) {
      for (int k = 1; k <= z; k++) {
        cntr = 0;
        int number = i * j * k;
        repeat[number]++;
        if (m.find(number) != m.end()) continue;
        for (int z = 1; z * z <= number; z++) {
          if (number % z == 0) {
            cntr++;
            if (z * z != number) cntr++;
          }
        }
        m[number] = cntr;
      }
    }
  }
  long long result = 0;
  for (auto it1 = m.begin(), it2 = repeat.begin(); it1 != m.end();
       it1++, it2++) {
    result += (it1->second % mod * it2->second % mod) % mod;
  }
  cout << result;
}
