#include <bits/stdc++.h>
using namespace std;
int main() {
  int a, b, c;
  cin >> a >> b >> c;
  long long sum = 0;
  const long long mod = 1073741824;
  vector<long long> dp(1000001, 0);
  for (long long i = 1; i <= a; i++)
    for (long long j = 1; j <= b; j++)
      for (long long k = 1; k <= c; k++)
        if (dp[i * j * k])
          sum = (sum + dp[i * j * k]) % mod;
        else {
          long long c = 0;
          for (int w = 1; w * w <= i * j * k; w++)
            if (i * j * k % w == 0)
              if (w == i * j * k / w)
                c++;
              else
                c += 2;
          dp[i * j * k] = c;
          sum = (sum + c) % mod;
        }
  cout << sum << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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;
  cin >> a >> b >> c;
  long long sum = 0;
  const long long mod = 1073741824;
  vector<long long> dp(1000001, 0);
  for (long long i = 1; i <= a; i++)
    for (long long j = 1; j <= b; j++)
      for (long long k = 1; k <= c; k++)
        if (dp[i * j * k])
          sum = (sum + dp[i * j * k]) % mod;
        else {
          long long c = 0;
          for (int w = 1; w * w <= i * j * k; w++)
            if (i * j * k % w == 0)
              if (w == i * j * k / w)
                c++;
              else
                c += 2;
          dp[i * j * k] = c;
          sum = (sum + c) % mod;
        }
  cout << sum << endl;
  return 0;
}

```