#include <bits/stdc++.h>
using namespace std;
vector<long long> seive(long long n) {
  vector<bool> is_prime(n + 1, 1);
  is_prime[0] = is_prime[1] = 0;
  for (long long i = 2; i * i <= n; ++i) {
    if (is_prime[i]) {
      for (long long j = i * i; j <= n; j += i) is_prime[j] = 0;
    }
  }
  vector<long long> prime;
  for (long long i = 0; i < n + 1; ++i)
    if (is_prime[i]) prime.push_back(i);
  return prime;
}
vector<long long> primes;
vector<long long> trial_division_with_primes(long long n) {
  vector<long long> factorization;
  for (long long d : primes) {
    if (d * d > n) break;
    while (n % d == 0) {
      factorization.push_back(d);
      n /= d;
    }
  }
  if (n > 1) factorization.push_back(n);
  return factorization;
}
long long int d(long long int n) {
  if (n <= 1) return 1;
  vector<long long int> factors = trial_division_with_primes(n);
  unordered_map<long long int, long long int> m;
  for (long long int i : factors) m[i]++;
  long long int res = 1;
  for (pair<long long int, long long int> x : m) {
    res *= ((x.second + 1) % 1073741824);
    res = res % 1073741824;
  }
  return res;
}
void solution() {
  long long int a, b, c;
  cin >> a >> b >> c;
  primes = seive((long long int)sqrt(a * b * c));
  vector<long long int> t(a * b * c + 1);
  for (long long int i = 0; i <= a * b * c; ++i) t[i] = d(i);
  long long int ans = 0;
  for (long long int i = 1; i <= a; ++i) {
    for (long long int j = 1; j <= b; ++j) {
      for (long long int k = 1; k <= c; ++k) {
        ans += (t[i * j * k] % 1073741824);
        ans = ans % 1073741824;
      }
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  unsigned long long tc = 1;
  for (unsigned long long t = 0; t < tc; ++t) solution();
  return 0;
}

### Prompt

Please create a solution in Cpp to 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;
vector<long long> seive(long long n) {
  vector<bool> is_prime(n + 1, 1);
  is_prime[0] = is_prime[1] = 0;
  for (long long i = 2; i * i <= n; ++i) {
    if (is_prime[i]) {
      for (long long j = i * i; j <= n; j += i) is_prime[j] = 0;
    }
  }
  vector<long long> prime;
  for (long long i = 0; i < n + 1; ++i)
    if (is_prime[i]) prime.push_back(i);
  return prime;
}
vector<long long> primes;
vector<long long> trial_division_with_primes(long long n) {
  vector<long long> factorization;
  for (long long d : primes) {
    if (d * d > n) break;
    while (n % d == 0) {
      factorization.push_back(d);
      n /= d;
    }
  }
  if (n > 1) factorization.push_back(n);
  return factorization;
}
long long int d(long long int n) {
  if (n <= 1) return 1;
  vector<long long int> factors = trial_division_with_primes(n);
  unordered_map<long long int, long long int> m;
  for (long long int i : factors) m[i]++;
  long long int res = 1;
  for (pair<long long int, long long int> x : m) {
    res *= ((x.second + 1) % 1073741824);
    res = res % 1073741824;
  }
  return res;
}
void solution() {
  long long int a, b, c;
  cin >> a >> b >> c;
  primes = seive((long long int)sqrt(a * b * c));
  vector<long long int> t(a * b * c + 1);
  for (long long int i = 0; i <= a * b * c; ++i) t[i] = d(i);
  long long int ans = 0;
  for (long long int i = 1; i <= a; ++i) {
    for (long long int j = 1; j <= b; ++j) {
      for (long long int k = 1; k <= c; ++k) {
        ans += (t[i * j * k] % 1073741824);
        ans = ans % 1073741824;
      }
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  unsigned long long tc = 1;
  for (unsigned long long t = 0; t < tc; ++t) solution();
  return 0;
}

```