#include <bits/stdc++.h>
using namespace std;
int prime[1005];
int miu[2005];
int h[2005];
int n, a, b, c;
int gcd(int a, int b) {
  if (b)
    return gcd(b, a % b);
  else
    return a;
}
void pre() {
  int i, j, c = 0;
  miu[1] = 1;
  for (i = 2; i <= n; ++i) {
    if (h[i] == 0) {
      prime[++c] = i;
      miu[i] = -1;
    }
    for (j = 1; j <= c && i * prime[j] <= n; ++j) {
      h[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        miu[i * prime[j]] = 0;
        break;
      }
      miu[i * prime[j]] = -miu[i];
    }
  }
}
int work() {
  int i, j, k, x, y;
  long long ans = 0;
  for (i = 1; i <= n; ++i)
    if (miu[i])
      for (j = 1; j <= n; ++j)
        if (gcd(i, j) == 1) {
          x = 0, y = 0;
          for (k = 1; k <= b / i; ++k)
            if (gcd(j, k) == 1) x += b / (k * i);
          for (k = 1; k <= c / i; ++k)
            if (gcd(j, k) == 1) y += c / (k * i);
          ans += (long long)miu[i] * (long long)x * (long long)y *
                     (long long)(a / j) +
                 1073741824;
          ans %= 1073741824;
        }
  return (ans + 1073741824) % 1073741824;
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  n = max(a, max(b, c));
  pre();
  printf("%d\n", work());
  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 prime[1005];
int miu[2005];
int h[2005];
int n, a, b, c;
int gcd(int a, int b) {
  if (b)
    return gcd(b, a % b);
  else
    return a;
}
void pre() {
  int i, j, c = 0;
  miu[1] = 1;
  for (i = 2; i <= n; ++i) {
    if (h[i] == 0) {
      prime[++c] = i;
      miu[i] = -1;
    }
    for (j = 1; j <= c && i * prime[j] <= n; ++j) {
      h[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        miu[i * prime[j]] = 0;
        break;
      }
      miu[i * prime[j]] = -miu[i];
    }
  }
}
int work() {
  int i, j, k, x, y;
  long long ans = 0;
  for (i = 1; i <= n; ++i)
    if (miu[i])
      for (j = 1; j <= n; ++j)
        if (gcd(i, j) == 1) {
          x = 0, y = 0;
          for (k = 1; k <= b / i; ++k)
            if (gcd(j, k) == 1) x += b / (k * i);
          for (k = 1; k <= c / i; ++k)
            if (gcd(j, k) == 1) y += c / (k * i);
          ans += (long long)miu[i] * (long long)x * (long long)y *
                     (long long)(a / j) +
                 1073741824;
          ans %= 1073741824;
        }
  return (ans + 1073741824) % 1073741824;
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  n = max(a, max(b, c));
  pre();
  printf("%d\n", work());
  return 0;
}

```