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:

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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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
unsigned int mu[10000 + 5], f[4000000 + 5], ans = 0;
int prime[10000 + 5], cnt = 0;
bool noprime[10000 + 5];
int main() {
  int a, b, c;
  scanf("%d%d%d", &a, &b, &c);
  ans = 0;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++)
      if (gcd(i, j) == 1) f[i * j] += (a / i) * (b / j);
  mu[1] = 1;
  for (int i = 2; i <= c; i++) {
    if (!noprime[i]) prime[++cnt] = i, mu[i] = -1;
    for (int j = 1; j <= cnt && i * prime[j] <= c; j++) {
      int now = i * prime[j];
      noprime[now] = 1;
      if (i % prime[j])
        mu[now] = -mu[i];
      else {
        mu[now] = 0;
        break;
      }
    }
  }
  for (int i = 1; i <= min(a * b, c); i++) {
    unsigned int sum1 = 0, sum2 = 0;
    for (int j = 1; i * j <= a * b; j++) sum1 += f[i * j];
    for (int j = 1; i * j <= c; j++) sum2 += c / (i * j);
    ans += sum1 * sum2 * mu[i];
  }
  printf("%d\n", ans & (1 << 30) - 1);
  return 0;
}