#include <bits/stdc++.h>
using namespace std;
const long long MAX = 1e6 + 6;
const long long inf = 1e9;
long long p[MAX];
long long d(long long x) {
  long long ans = 1;
  while (x > 1) {
    long long fac = p[x], e = 0;
    while (x % fac == 0) x /= fac, e++;
    ans *= (e + 1);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  for (long long i = 2; i < (long long)MAX; i++) {
    if (p[i] == 0)
      for (long long j = 1; j < (long long)MAX / i; j++) p[i * j] = i;
  }
  long long ans = 0, a, b, c;
  cin >> a >> b >> c;
  for (long long i = 1; i < (long long)a + 1; i++)
    for (long long j = 1; j < (long long)b + 1; j++)
      for (long long k = 1; k < (long long)c + 1; k++) ans += d(i * j * k);
  cout << ans % 1073741824 << endl;
}

### Prompt

In Cpp, your task is to solve 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;
const long long MAX = 1e6 + 6;
const long long inf = 1e9;
long long p[MAX];
long long d(long long x) {
  long long ans = 1;
  while (x > 1) {
    long long fac = p[x], e = 0;
    while (x % fac == 0) x /= fac, e++;
    ans *= (e + 1);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  for (long long i = 2; i < (long long)MAX; i++) {
    if (p[i] == 0)
      for (long long j = 1; j < (long long)MAX / i; j++) p[i * j] = i;
  }
  long long ans = 0, a, b, c;
  cin >> a >> b >> c;
  for (long long i = 1; i < (long long)a + 1; i++)
    for (long long j = 1; j < (long long)b + 1; j++)
      for (long long k = 1; k < (long long)c + 1; k++) ans += d(i * j * k);
  cout << ans % 1073741824 << endl;
}

```