#include <bits/stdc++.h>
using namespace std;
inline int read(register int ans = 0, register int sgn = ' ',
                register int ch = getchar()) {
  for (; ch < '0' || ch > '9'; sgn = ch, ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; (ans *= 10) += ch - '0', ch = getchar())
    ;
  return sgn - '-' ? ans : -ans;
}
bitset<2000 + 5> book;
int P[2000 + 5], mu[2000 + 5], tot, a, b, c, lim;
long long gcd[2000 + 5][2000 + 5], ans;
inline long long GCD(long long a, long long b) {
  if (!b) return gcd[a][b] = gcd[b][a] = a;
  if (gcd[a][b]) return gcd[a][b];
  return gcd[a][b] = gcd[b][a] = GCD(b, a % b);
}
void Init() {
  mu[0] = 0, mu[1] = 1;
  for (register long long i = 2, j; i <= 2000; i++)
    for (((bool)book[i] ?: (P[tot++] = i, mu[i] = -1)), j = 0;
         j < tot && i * P[j] <= 2000; book[i * P[j]] = 1, j++)
      if (i % P[j])
        mu[i * P[j]] = -mu[i];
      else {
        mu[i * P[j]] = 0, book[i * P[j]] = 1;
        break;
      }
  return;
}
inline long long Calc(int d, int n, long long k = 0) {
  for (register int i = 1; i <= d;
       (GCD(i, n) == 1 ? k = (k + d / i) % 1073741824 : 0), i++)
    ;
  return k;
}
int main() {
  a = read(), b = read(), c = read(), Init(), lim = min(b, c);
  for (register int i = 1, j; i <= a; i++)
    for (j = 1; j <= lim; j++)
      if (GCD(i, j) == 1)
        ans = (ans + 1ll * (a / i) * mu[j] * Calc(b / j, i) * Calc(c / j, i) %
                         1073741824) %
              1073741824;
  printf("%lld\n", ans);
  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>
using namespace std;
inline int read(register int ans = 0, register int sgn = ' ',
                register int ch = getchar()) {
  for (; ch < '0' || ch > '9'; sgn = ch, ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; (ans *= 10) += ch - '0', ch = getchar())
    ;
  return sgn - '-' ? ans : -ans;
}
bitset<2000 + 5> book;
int P[2000 + 5], mu[2000 + 5], tot, a, b, c, lim;
long long gcd[2000 + 5][2000 + 5], ans;
inline long long GCD(long long a, long long b) {
  if (!b) return gcd[a][b] = gcd[b][a] = a;
  if (gcd[a][b]) return gcd[a][b];
  return gcd[a][b] = gcd[b][a] = GCD(b, a % b);
}
void Init() {
  mu[0] = 0, mu[1] = 1;
  for (register long long i = 2, j; i <= 2000; i++)
    for (((bool)book[i] ?: (P[tot++] = i, mu[i] = -1)), j = 0;
         j < tot && i * P[j] <= 2000; book[i * P[j]] = 1, j++)
      if (i % P[j])
        mu[i * P[j]] = -mu[i];
      else {
        mu[i * P[j]] = 0, book[i * P[j]] = 1;
        break;
      }
  return;
}
inline long long Calc(int d, int n, long long k = 0) {
  for (register int i = 1; i <= d;
       (GCD(i, n) == 1 ? k = (k + d / i) % 1073741824 : 0), i++)
    ;
  return k;
}
int main() {
  a = read(), b = read(), c = read(), Init(), lim = min(b, c);
  for (register int i = 1, j; i <= a; i++)
    for (j = 1; j <= lim; j++)
      if (GCD(i, j) == 1)
        ans = (ans + 1ll * (a / i) * mu[j] * Calc(b / j, i) * Calc(c / j, i) %
                         1073741824) %
              1073741824;
  printf("%lld\n", ans);
  return 0;
}

```