#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 22;
const long long M = 1073741824;
int n;
map<int, int> m;
long long f(int x) {
  if (m.count(x)) return m[x];
  int t = x;
  long long r = 1;
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      int cnt = 1;
      while (x % i == 0) {
        x /= i;
        cnt++;
      }
      r *= cnt;
      r %= M;
    }
  }
  if (x > 1) r *= 2;
  r %= M;
  return m[t] = r;
}
int main() {
  int a, b, c;
  long long sum = 0;
  cin >> a >> b >> c;
  for (int i = 1; i <= (a); i++)
    for (int j = 1; j <= (b); j++)
      for (int k = 1; k <= (c); k++) sum = (sum + f(i * j * k)) % M;
  cout << sum;
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
const int N = 1e5 + 22;
const long long M = 1073741824;
int n;
map<int, int> m;
long long f(int x) {
  if (m.count(x)) return m[x];
  int t = x;
  long long r = 1;
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      int cnt = 1;
      while (x % i == 0) {
        x /= i;
        cnt++;
      }
      r *= cnt;
      r %= M;
    }
  }
  if (x > 1) r *= 2;
  r %= M;
  return m[t] = r;
}
int main() {
  int a, b, c;
  long long sum = 0;
  cin >> a >> b >> c;
  for (int i = 1; i <= (a); i++)
    for (int j = 1; j <= (b); j++)
      for (int k = 1; k <= (c); k++) sum = (sum + f(i * j * k)) % M;
  cout << sum;
  return 0;
}

```