#include <bits/stdc++.h>
using namespace std;
int gcd(int x, int y) { return x ? gcd(y % x, x) : y; }
bool sqf(int x) {
  int i;
  for (i = 2; i * i <= x; i++)
    if (x % (i * i) == 0) return false;
  return true;
}
int sgn(int x) {
  int i;
  int ans = 1;
  for (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, i;
  for (i = 1; i <= M; i++)
    if (gcd(x, i) == 1) ans += M / i;
  return ans;
}
int main(void) {
  int A, B, C, s, x;
  cin >> A >> B >> C;
  long long ans = 0;
  for (s = 1; s <= 2000; s++)
    if (sqf(s)) {
      int coef = sgn(s);
      for (x = 1; x <= A; x++)
        if (gcd(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) << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 gcd(int x, int y) { return x ? gcd(y % x, x) : y; }
bool sqf(int x) {
  int i;
  for (i = 2; i * i <= x; i++)
    if (x % (i * i) == 0) return false;
  return true;
}
int sgn(int x) {
  int i;
  int ans = 1;
  for (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, i;
  for (i = 1; i <= M; i++)
    if (gcd(x, i) == 1) ans += M / i;
  return ans;
}
int main(void) {
  int A, B, C, s, x;
  cin >> A >> B >> C;
  long long ans = 0;
  for (s = 1; s <= 2000; s++)
    if (sqf(s)) {
      int coef = sgn(s);
      for (x = 1; x <= A; x++)
        if (gcd(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) << endl;
  return 0;
}

```