#include <bits/stdc++.h>
using namespace std;
const int mod = 1073741824;
int mu[2010], pri[2010], tot;
bool check[2010];
void get_mu(int n) {
  mu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!check[i]) pri[++tot] = i, mu[i] = -1;
    for (int j = 1; j <= tot && i * pri[j] <= n; j++) {
      check[i * pri[j]] = true;
      if (i % pri[j] == 0) break;
      mu[i * pri[j]] = mu[i] * mu[pri[j]];
    }
  }
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int sum(int n, int x) {
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (gcd(i, x) == 1) cnt += n / i;
  return cnt;
}
int main() {
  int n, m, K;
  scanf("%d%d%d", &n, &m, &K);
  get_mu(m);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int d = 1; d <= m; d++) {
      if (gcd(i, d) != 1) continue;
      ans += (n / i) * mu[d] * 1LL * sum(m / d, i) % mod * 1LL * sum(K / d, i) %
             mod;
      if (ans >= mod) ans -= mod;
    }
  }
  printf("%d\n", ans);
  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;
const int mod = 1073741824;
int mu[2010], pri[2010], tot;
bool check[2010];
void get_mu(int n) {
  mu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!check[i]) pri[++tot] = i, mu[i] = -1;
    for (int j = 1; j <= tot && i * pri[j] <= n; j++) {
      check[i * pri[j]] = true;
      if (i % pri[j] == 0) break;
      mu[i * pri[j]] = mu[i] * mu[pri[j]];
    }
  }
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int sum(int n, int x) {
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (gcd(i, x) == 1) cnt += n / i;
  return cnt;
}
int main() {
  int n, m, K;
  scanf("%d%d%d", &n, &m, &K);
  get_mu(m);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int d = 1; d <= m; d++) {
      if (gcd(i, d) != 1) continue;
      ans += (n / i) * mu[d] * 1LL * sum(m / d, i) % mod * 1LL * sum(K / d, i) %
             mod;
      if (ans >= mod) ans -= mod;
    }
  }
  printf("%d\n", ans);
  return 0;
}

```