Problem: 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 ≤ 2000).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Examples

Input

2 2 2


Output

20


Input

4 4 4


Output

328


Input

10 10 10


Output

11536

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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void print(vector<int> &v) {
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << endl;
}
void div(int a, vector<int> &v) {
  int ans = 0;
  for (int i = 1; i <= sqrt(a); i++) {
    if (a % i == 0) {
      if (a / i == i) {
        ans++;
      } else {
        ans += 2;
      }
    }
  }
  v[a] = ans;
}
int main() {
  int a, b, c;
  cin >> a >> b >> c;
  vector<int> v(1000005, -1);
  long long int ans = 0;
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        if (v[i * j * k] == -1) {
          div(i * j * k, v);
          ans += v[i * j * k];
        } else {
          ans += v[i * j * k];
        }
      }
    }
  }
  cout << ans % 1073741824 << endl;
}