#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
bool isprime(long long n) {
  if (n == 0 || n == 1) return 0;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return 0;
    }
  }
  return 1;
}
int mod = (1 << 30);
int main() {
  vector<int> div(1000000);
  for (int i = 1; i <= 1000000; i++) {
    for (int j = i; j <= 1000000; j += i) div[j]++;
  }
  int a, b, c, d = 0;
  cin >> a >> b >> c;
  for (long long i = 1; i <= a; i++) {
    for (long long j = 1; j <= b; j++) {
      for (long long k = 1; k <= c; k++) {
        d += div[i * j * k];
        d %= mod;
      }
    }
  }
  cout << d;
  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 gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
bool isprime(long long n) {
  if (n == 0 || n == 1) return 0;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return 0;
    }
  }
  return 1;
}
int mod = (1 << 30);
int main() {
  vector<int> div(1000000);
  for (int i = 1; i <= 1000000; i++) {
    for (int j = i; j <= 1000000; j += i) div[j]++;
  }
  int a, b, c, d = 0;
  cin >> a >> b >> c;
  for (long long i = 1; i <= a; i++) {
    for (long long j = 1; j <= b; j++) {
      for (long long k = 1; k <= c; k++) {
        d += div[i * j * k];
        d %= mod;
      }
    }
  }
  cout << d;
  return 0;
}

```