#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
const int mod = 1 << 30;
int gc[N][N], mu[N], prime[N];
bool vis[N];
int cnt;
void get_prime(int m) {
  mu[1] = 1;
  for (int i = 2; i <= m; i++) {
    if (!vis[i]) {
      mu[i] = -1;
      prime[++cnt] = i;
    }
    for (int j = 1; j <= cnt && prime[j] * i <= m; j++) {
      vis[prime[j] * i] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      } else
        mu[i * prime[j]] = -mu[i];
    }
  }
}
int gcd(int a, int b) {
  if (gc[a][b]) return gc[a][b];
  if (b == 0) return a;
  return gc[a][b] = gcd(b, a % b);
}
int a, b, c;
long long cal(int d, int x) {
  long long ans = 0;
  for (int i = 1; i <= d; i++) {
    if (gcd(i, x) == 1) ans += (long long)(d / i);
  }
  return ans;
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  get_prime(2000);
  long long ans = 0;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= min(b, c); j++) {
      if (gcd(i, j) == 1)
        ans = (ans % mod + (long long)(a / i) * mu[j] * cal(b / j, i) *
                               cal(c / j, i) % mod) %
              mod;
    }
  cout << ans;
}

### 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 = 2005;
const int mod = 1 << 30;
int gc[N][N], mu[N], prime[N];
bool vis[N];
int cnt;
void get_prime(int m) {
  mu[1] = 1;
  for (int i = 2; i <= m; i++) {
    if (!vis[i]) {
      mu[i] = -1;
      prime[++cnt] = i;
    }
    for (int j = 1; j <= cnt && prime[j] * i <= m; j++) {
      vis[prime[j] * i] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      } else
        mu[i * prime[j]] = -mu[i];
    }
  }
}
int gcd(int a, int b) {
  if (gc[a][b]) return gc[a][b];
  if (b == 0) return a;
  return gc[a][b] = gcd(b, a % b);
}
int a, b, c;
long long cal(int d, int x) {
  long long ans = 0;
  for (int i = 1; i <= d; i++) {
    if (gcd(i, x) == 1) ans += (long long)(d / i);
  }
  return ans;
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  get_prime(2000);
  long long ans = 0;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= min(b, c); j++) {
      if (gcd(i, j) == 1)
        ans = (ans % mod + (long long)(a / i) * mu[j] * cal(b / j, i) *
                               cal(c / j, i) % mod) %
              mod;
    }
  cout << ans;
}

```