#include <bits/stdc++.h>
using namespace std;
int x, y, z, cnt = 0, cnt1 = 0, i, arr[1000005];
void divisors(int x) {
  cnt = 0;
  for (i = 1; i * i < x; i++) {
    if (!(x % i)) {
      cnt += 2;
    }
  }
  if (i * i == x) cnt++;
  arr[x] = cnt;
  return;
}
int main() {
  cin >> x >> y >> z;
  for (int i = 1; i <= x; i++) {
    for (int j = 1; j <= y; j++) {
      for (int k = 1; k <= z; k++) {
        if (arr[i * j * k] == 0) {
          divisors(i * j * k);
          cnt1 += arr[i * j * k];
        } else
          cnt1 += arr[i * j * k];
      }
    }
  }
  cout << cnt1 << endl;
  return 0;
}

### Prompt

Generate 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 x, y, z, cnt = 0, cnt1 = 0, i, arr[1000005];
void divisors(int x) {
  cnt = 0;
  for (i = 1; i * i < x; i++) {
    if (!(x % i)) {
      cnt += 2;
    }
  }
  if (i * i == x) cnt++;
  arr[x] = cnt;
  return;
}
int main() {
  cin >> x >> y >> z;
  for (int i = 1; i <= x; i++) {
    for (int j = 1; j <= y; j++) {
      for (int k = 1; k <= z; k++) {
        if (arr[i * j * k] == 0) {
          divisors(i * j * k);
          cnt1 += arr[i * j * k];
        } else
          cnt1 += arr[i * j * k];
      }
    }
  }
  cout << cnt1 << endl;
  return 0;
}

```