#include <bits/stdc++.h>
using namespace std;
int N = 2000, a, b, c, M[3010], modulo = 1 << 30, ans;
int gcd(int x, int y) {
  for (int t; y; t = x % y, x = y, y = t)
    ;
  return x;
}
int getsum(int x, int y) {
  int sum = 0;
  for (int i = 1; i <= x; i++)
    if (gcd(i, y) == 1) sum += x / i;
  return sum;
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  M[1] = 1;
  for (int i = 2; i <= N; i++)
    for (int j = 1; j < i; j++)
      if (i % j == 0) M[i] -= M[j];
  ans = 0;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= N; j++)
      if (gcd(i, j) == 1)
        ans += M[j] * (a / i) * getsum(b / j, i) * getsum(c / j, i);
  printf("%d\n", (ans % modulo + modulo) % modulo);
  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;
int N = 2000, a, b, c, M[3010], modulo = 1 << 30, ans;
int gcd(int x, int y) {
  for (int t; y; t = x % y, x = y, y = t)
    ;
  return x;
}
int getsum(int x, int y) {
  int sum = 0;
  for (int i = 1; i <= x; i++)
    if (gcd(i, y) == 1) sum += x / i;
  return sum;
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  M[1] = 1;
  for (int i = 2; i <= N; i++)
    for (int j = 1; j < i; j++)
      if (i % j == 0) M[i] -= M[j];
  ans = 0;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= N; j++)
      if (gcd(i, j) == 1)
        ans += M[j] * (a / i) * getsum(b / j, i) * getsum(c / j, i);
  printf("%d\n", (ans % modulo + modulo) % modulo);
  return 0;
}

```