#include <bits/stdc++.h>
using namespace std;
map<int, int> res;
int arr[1000010] = {0};
long long cal(long long num) {
  if (arr[num] > 0) return arr[num];
  for (long long i = 1; i * i <= num; i++) {
    if (num % i == 0) {
      if (i * i == num)
        arr[num] += 1;
      else
        arr[num] += 2;
    }
  }
  return arr[num];
}
int main() {
  int a, b, c;
  arr[1] = 1;
  arr[2] = 2;
  long long ans = 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++) {
        long long tmp = i * j * k;
        ans += cal(tmp);
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
map<int, int> res;
int arr[1000010] = {0};
long long cal(long long num) {
  if (arr[num] > 0) return arr[num];
  for (long long i = 1; i * i <= num; i++) {
    if (num % i == 0) {
      if (i * i == num)
        arr[num] += 1;
      else
        arr[num] += 2;
    }
  }
  return arr[num];
}
int main() {
  int a, b, c;
  arr[1] = 1;
  arr[2] = 2;
  long long ans = 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++) {
        long long tmp = i * j * k;
        ans += cal(tmp);
      }
    }
  }
  cout << ans;
  return 0;
}

```