#include <bits/stdc++.h>
using namespace std;
int nwd(int a, int b) { return !b ? a : nwd(b, a % b); }
bool sqf(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % (i * i) == 0) return false;
  return true;
}
int sgn(int x) {
  int ans = 1;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) ans = -ans, x /= i;
  if (x > 1) ans = -ans;
  return ans;
}
int func(int x, int M) {
  int ans = 0;
  for (int i = 1; i <= M; i++)
    if (nwd(x, i) == 1) ans += M / i;
  return ans;
}
int a, b, c;
long long ans;
int main() {
  scanf("%d %d %d", &a, &b, &c);
  for (int s = 1; s <= 2000; s++)
    if (sqf(s)) {
      int coef = sgn(s);
      for (int x = 1; x <= a; x++)
        if (nwd(x, s) == 1)
          ans += (long long)(a / x) * (long long)coef *
                 (long long)func(x, b / s) * (long long)func(x, c / s);
    }
  cout << ans % (1 << 30) << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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 nwd(int a, int b) { return !b ? a : nwd(b, a % b); }
bool sqf(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % (i * i) == 0) return false;
  return true;
}
int sgn(int x) {
  int ans = 1;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) ans = -ans, x /= i;
  if (x > 1) ans = -ans;
  return ans;
}
int func(int x, int M) {
  int ans = 0;
  for (int i = 1; i <= M; i++)
    if (nwd(x, i) == 1) ans += M / i;
  return ans;
}
int a, b, c;
long long ans;
int main() {
  scanf("%d %d %d", &a, &b, &c);
  for (int s = 1; s <= 2000; s++)
    if (sqf(s)) {
      int coef = sgn(s);
      for (int x = 1; x <= a; x++)
        if (nwd(x, s) == 1)
          ans += (long long)(a / x) * (long long)coef *
                 (long long)func(x, b / s) * (long long)func(x, c / s);
    }
  cout << ans % (1 << 30) << "\n";
  return 0;
}

```