#include <bits/stdc++.h>
using namespace std;
unsigned int a, b, c, Kind[2003][2003], ans, K[2003], Mk[2003];
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int Get(int n, int m) {
  int res = 0;
  for (int i = 1; i <= n; i++)
    if (Kind[i][m]) res += n / i;
  return res;
}
int main() {
  for (int i = 1; i != 2003; ++i) K[i] = 1;
  for (int i = 2; i != 2003; ++i) {
    if (Mk[i]) continue;
    for (int j = i; j < 2003; j += i) {
      ((j / i % i == 0) && (K[j] = 0));
      K[j] = -K[j], Mk[j] = 1;
    }
  }
  scanf("%u %u %u", &a, &b, &c);
  if (a > b) swap(a, b);
  if (b > c) swap(b, c);
  if (a > b) swap(a, b);
  for (int i = 1; i <= c; i++)
    for (int j = 1; j <= c; j++) Kind[i][j] = gcd(i, j) == 1;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= min(b, c); j++)
      if (Kind[i][j]) ans += K[j] * (a / i) * Get(b / j, i) * Get(c / j, i);
  ans <<= 2, ans >>= 2;
  printf("%u\n", ans);
  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;
unsigned int a, b, c, Kind[2003][2003], ans, K[2003], Mk[2003];
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int Get(int n, int m) {
  int res = 0;
  for (int i = 1; i <= n; i++)
    if (Kind[i][m]) res += n / i;
  return res;
}
int main() {
  for (int i = 1; i != 2003; ++i) K[i] = 1;
  for (int i = 2; i != 2003; ++i) {
    if (Mk[i]) continue;
    for (int j = i; j < 2003; j += i) {
      ((j / i % i == 0) && (K[j] = 0));
      K[j] = -K[j], Mk[j] = 1;
    }
  }
  scanf("%u %u %u", &a, &b, &c);
  if (a > b) swap(a, b);
  if (b > c) swap(b, c);
  if (a > b) swap(a, b);
  for (int i = 1; i <= c; i++)
    for (int j = 1; j <= c; j++) Kind[i][j] = gcd(i, j) == 1;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= min(b, c); j++)
      if (Kind[i][j]) ans += K[j] * (a / i) * Get(b / j, i) * Get(c / j, i);
  ans <<= 2, ans >>= 2;
  printf("%u\n", ans);
  return 0;
}

```