#include <bits/stdc++.h>
using namespace std;
int sumDivisors(int num) {
  int numberOfDivisors = 1;
  for (int i = 2; i <= num; i++) {
    int exponent = 0;
    while (num % i == 0) {
      exponent++;
      num /= i;
    }
    numberOfDivisors *= (exponent + 1);
  }
  return numberOfDivisors;
}
int main() {
  int a, b, c;
  cin >> a >> b >> c;
  long int sum = 0;
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        sum += sumDivisors(i * j * k);
      }
    }
  }
  cout << sum % 1073741824;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 sumDivisors(int num) {
  int numberOfDivisors = 1;
  for (int i = 2; i <= num; i++) {
    int exponent = 0;
    while (num % i == 0) {
      exponent++;
      num /= i;
    }
    numberOfDivisors *= (exponent + 1);
  }
  return numberOfDivisors;
}
int main() {
  int a, b, c;
  cin >> a >> b >> c;
  long int sum = 0;
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        sum += sumDivisors(i * j * k);
      }
    }
  }
  cout << sum % 1073741824;
  return 0;
}

```