#include <bits/stdc++.h>
using namespace std;
int u[2002];
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long sumcop(int a, int b) {
  long long ans = 0;
  for (int i = 1; i <= a; i++) {
    if (gcd(i, b) == 1) ans = (ans + (a / i)) % 1073741824;
  }
  return ans;
}
int main() {
  u[1] = 1;
  for (int i = 1; i < 2002; i++) {
    for (int j = 2 * i; j < 2002; j += i) u[j] -= u[i];
  }
  long long ans = 0;
  int a, b, c;
  scanf("%d %d %d", &a, &b, &c);
  for (int i = 1; i <= a; i++) {
    for (int r = 1; r <= max(b, c); r++) {
      if (gcd(r, i) == 1) {
        ans = (ans + (a / i) * u[r] * sumcop(b / r, i) * sumcop(c / r, i)) %
              1073741824;
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please formulate 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 u[2002];
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long sumcop(int a, int b) {
  long long ans = 0;
  for (int i = 1; i <= a; i++) {
    if (gcd(i, b) == 1) ans = (ans + (a / i)) % 1073741824;
  }
  return ans;
}
int main() {
  u[1] = 1;
  for (int i = 1; i < 2002; i++) {
    for (int j = 2 * i; j < 2002; j += i) u[j] -= u[i];
  }
  long long ans = 0;
  int a, b, c;
  scanf("%d %d %d", &a, &b, &c);
  for (int i = 1; i <= a; i++) {
    for (int r = 1; r <= max(b, c); r++) {
      if (gcd(r, i) == 1) {
        ans = (ans + (a / i) * u[r] * sumcop(b / r, i) * sumcop(c / r, i)) %
              1073741824;
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```