#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b, c, i, j, k, x;
  cin >> a >> b >> c;
  long long int mul, sum = 0;
  vector<int> v(1000000, 0);
  for (i = 1; i <= a; i = i + 1) {
    for (j = 1; j <= b; j = j + 1) {
      for (k = 1; k <= c; k = k + 1) {
        mul = i * j * k;
        if (v[mul] == 0) {
          int e = 0;
          for (x = 1; x * x <= mul; x = x + 1) {
            if (mul % x == 0) {
              if (x == mul / x) {
                e = e + 1;
                sum = sum + 1;
              } else {
                e = e + 2;
                sum = sum + 2;
              }
            }
          }
          v[mul] = e;
        } else {
          sum = sum + v[mul];
        }
        if (sum >= 1073741824) {
          sum = sum % 1073741824;
        }
      }
    }
  }
  cout << sum << endl;
}

### Prompt

Your challenge is to write a cpp solution to the following 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:

<image>

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b, c, i, j, k, x;
  cin >> a >> b >> c;
  long long int mul, sum = 0;
  vector<int> v(1000000, 0);
  for (i = 1; i <= a; i = i + 1) {
    for (j = 1; j <= b; j = j + 1) {
      for (k = 1; k <= c; k = k + 1) {
        mul = i * j * k;
        if (v[mul] == 0) {
          int e = 0;
          for (x = 1; x * x <= mul; x = x + 1) {
            if (mul % x == 0) {
              if (x == mul / x) {
                e = e + 1;
                sum = sum + 1;
              } else {
                e = e + 2;
                sum = sum + 2;
              }
            }
          }
          v[mul] = e;
        } else {
          sum = sum + v[mul];
        }
        if (sum >= 1073741824) {
          sum = sum % 1073741824;
        }
      }
    }
  }
  cout << sum << endl;
}

```