#include <bits/stdc++.h>
const int MOD = 1073741824;
const int N = 2000 + 5;
const int dx[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};
using namespace std;
int mu[N];
int prime[N];
bool bprime[N];
void getMu(int n) {
  mu[1] = 1;
  int cnt = 0;
  memset(bprime, false, sizeof(bprime));
  for (int i = 2; i <= n; i++) {
    if (!bprime[i]) {
      prime[cnt++] = i;
      mu[i] = -1;
    }
    for (int j = 0; j < cnt; j++) {
      int k = i * prime[j];
      if (k > n) break;
      bprime[k] = true;
      if (i % prime[j])
        mu[k] = -mu[i];
      else {
        mu[k] = 0;
        break;
      }
    }
  }
}
int a, b, c;
int table[N][N];
void minSort() {
  if (a > b) swap(a, b);
  if (b > c) swap(b, c);
  if (a > b) swap(a, b);
}
int GCD(int x, int y) { return y ? GCD(y, x % y) : x; }
int F(int n, int m) {
  int res = 0;
  for (int i = 1; i <= n; i++)
    if (table[i][m] == 1) res += n / i;
  return res;
}
int main() {
  getMu(N);
  scanf("%d%d%d", &a, &b, &c);
  minSort();
  for (int i = 1; i <= c; i++) {
    for (int j = i; j <= c; j++) {
      table[i][j] = GCD(i, j);
      table[j][i] = GCD(i, j);
    }
  }
  int res = 0;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= min(b, c); j++)
      if (table[i][j] == 1) res += mu[j] * (a / i) * F(b / j, i) * F(c / j, i);
  res = (res % MOD + MOD) % MOD;
  printf("%d\n", res);
  return 0;
}

### Prompt

Your challenge is to write 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>
const int MOD = 1073741824;
const int N = 2000 + 5;
const int dx[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};
using namespace std;
int mu[N];
int prime[N];
bool bprime[N];
void getMu(int n) {
  mu[1] = 1;
  int cnt = 0;
  memset(bprime, false, sizeof(bprime));
  for (int i = 2; i <= n; i++) {
    if (!bprime[i]) {
      prime[cnt++] = i;
      mu[i] = -1;
    }
    for (int j = 0; j < cnt; j++) {
      int k = i * prime[j];
      if (k > n) break;
      bprime[k] = true;
      if (i % prime[j])
        mu[k] = -mu[i];
      else {
        mu[k] = 0;
        break;
      }
    }
  }
}
int a, b, c;
int table[N][N];
void minSort() {
  if (a > b) swap(a, b);
  if (b > c) swap(b, c);
  if (a > b) swap(a, b);
}
int GCD(int x, int y) { return y ? GCD(y, x % y) : x; }
int F(int n, int m) {
  int res = 0;
  for (int i = 1; i <= n; i++)
    if (table[i][m] == 1) res += n / i;
  return res;
}
int main() {
  getMu(N);
  scanf("%d%d%d", &a, &b, &c);
  minSort();
  for (int i = 1; i <= c; i++) {
    for (int j = i; j <= c; j++) {
      table[i][j] = GCD(i, j);
      table[j][i] = GCD(i, j);
    }
  }
  int res = 0;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= min(b, c); j++)
      if (table[i][j] == 1) res += mu[j] * (a / i) * F(b / j, i) * F(c / j, i);
  res = (res % MOD + MOD) % MOD;
  printf("%d\n", res);
  return 0;
}

```