#include <bits/stdc++.h>
using namespace std;
int a, b, c, ans;
int p[2005], prime[2005], mu[2005], G[2005][2005];
void get(int n) {
  mu[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!p[i]) {
      prime[++prime[0]] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= prime[0] && i * prime[j] <= n; ++j) {
      p[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      } else
        mu[i * prime[j]] = -mu[i];
    }
  }
}
void Min(int &a, int &b, int &c) {
  if (a > b) swap(a, b);
  if (b > c) swap(b, c);
  if (a > b) swap(a, b);
}
int gcd(int a, int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
int F(int n, int m) {
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    if (G[i][m] == 1) ans += n / i;
  return ans;
}
int main() {
  get(2000);
  scanf("%d%d%d", &a, &b, &c);
  Min(a, b, c);
  for (int i = 1; i <= c; ++i)
    for (int j = i; j <= c; ++j) G[i][j] = G[j][i] = gcd(i, j);
  for (int i = 1; i <= a; ++i)
    for (int j = 1; j <= min(b, c); ++j)
      if (G[i][j] == 1) ans += mu[j] * (a / i) * F(b / j, i) * F(c / j, i);
  ans = (ans % 1073741824 + 1073741824) % 1073741824;
  printf("%d\n", ans);
}

### Prompt

Generate a Cpp solution 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;
int a, b, c, ans;
int p[2005], prime[2005], mu[2005], G[2005][2005];
void get(int n) {
  mu[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!p[i]) {
      prime[++prime[0]] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= prime[0] && i * prime[j] <= n; ++j) {
      p[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      } else
        mu[i * prime[j]] = -mu[i];
    }
  }
}
void Min(int &a, int &b, int &c) {
  if (a > b) swap(a, b);
  if (b > c) swap(b, c);
  if (a > b) swap(a, b);
}
int gcd(int a, int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
int F(int n, int m) {
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    if (G[i][m] == 1) ans += n / i;
  return ans;
}
int main() {
  get(2000);
  scanf("%d%d%d", &a, &b, &c);
  Min(a, b, c);
  for (int i = 1; i <= c; ++i)
    for (int j = i; j <= c; ++j) G[i][j] = G[j][i] = gcd(i, j);
  for (int i = 1; i <= a; ++i)
    for (int j = 1; j <= min(b, c); ++j)
      if (G[i][j] == 1) ans += mu[j] * (a / i) * F(b / j, i) * F(c / j, i);
  ans = (ans % 1073741824 + 1073741824) % 1073741824;
  printf("%d\n", ans);
}

```