#include <bits/stdc++.h>
using namespace std;
const int N = 2003, MOD = 1073741824;
int mu[N + 3], prime[N + 3], pd[N + 3], n, m, t, g[N][N];
void init() {
  mu[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (!pd[i]) {
      prime[++prime[0]] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= prime[0]; j++) {
      if (i * prime[j] > N) break;
      pd[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
  }
}
int gcd(int x, int y) {
  if (g[x][y]) return g[x][y];
  int r, X = x, Y = y;
  while (y) {
    r = x % y;
    x = y;
    y = r;
  }
  g[X][Y] = x;
  return x;
}
int calc(int n, int d, int k) {
  int ans = 0;
  for (int i = 1; i <= n; i++)
    if (gcd(d * i, k) == 1) ans += (n / i);
  return ans;
}
int main() {
  scanf("%d%d%d", &n, &m, &t);
  if (n > m) swap(n, m);
  if (n > t) swap(n, t);
  init();
  int ans = 0;
  for (int k = 1; k <= t; k++)
    for (int d = 1; d <= n; d++)
      ans += mu[d] * (t / k) * calc(n / d, d, k) * calc(m / d, d, k);
  printf("%d\n", (ans % MOD + MOD) % MOD);
}

### Prompt

In cpp, your task is to solve 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 N = 2003, MOD = 1073741824;
int mu[N + 3], prime[N + 3], pd[N + 3], n, m, t, g[N][N];
void init() {
  mu[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (!pd[i]) {
      prime[++prime[0]] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= prime[0]; j++) {
      if (i * prime[j] > N) break;
      pd[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
  }
}
int gcd(int x, int y) {
  if (g[x][y]) return g[x][y];
  int r, X = x, Y = y;
  while (y) {
    r = x % y;
    x = y;
    y = r;
  }
  g[X][Y] = x;
  return x;
}
int calc(int n, int d, int k) {
  int ans = 0;
  for (int i = 1; i <= n; i++)
    if (gcd(d * i, k) == 1) ans += (n / i);
  return ans;
}
int main() {
  scanf("%d%d%d", &n, &m, &t);
  if (n > m) swap(n, m);
  if (n > t) swap(n, t);
  init();
  int ans = 0;
  for (int k = 1; k <= t; k++)
    for (int d = 1; d <= n; d++)
      ans += mu[d] * (t / k) * calc(n / d, d, k) * calc(m / d, d, k);
  printf("%d\n", (ans % MOD + MOD) % MOD);
}

```