#include <bits/stdc++.h>
using namespace std;
long long int divs[1000001];
int main() {
  long long int a, b, c, i, j, k, no, sum = 0, d;
  scanf("%lld%lld%lld", &a, &b, &c);
  for (i = 1; i <= 1000000; i++) {
    for (j = i; j <= 1000000; j += i) {
      divs[j]++;
    }
  }
  for (i = 1; i <= a; i++) {
    for (j = 1; j <= b; j++) {
      for (k = 1; k <= c; k++) {
        no = i * j * k;
        d = divs[no];
        sum = (sum + d) % 1073741824;
      }
    }
  }
  printf("%lld", sum);
}

### 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;
long long int divs[1000001];
int main() {
  long long int a, b, c, i, j, k, no, sum = 0, d;
  scanf("%lld%lld%lld", &a, &b, &c);
  for (i = 1; i <= 1000000; i++) {
    for (j = i; j <= 1000000; j += i) {
      divs[j]++;
    }
  }
  for (i = 1; i <= a; i++) {
    for (j = 1; j <= b; j++) {
      for (k = 1; k <= c; k++) {
        no = i * j * k;
        d = divs[no];
        sum = (sum + d) % 1073741824;
      }
    }
  }
  printf("%lld", sum);
}

```