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 read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
unsigned int mu[2010], f[2010][2010];
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline unsigned int solve(int a, int b) {
  if (f[a][b]) return f[a][b];
  unsigned int& res = f[a][b];
  for (int i = (1); i <= (b); i++)
    if (gcd(a, i) == 1) res += b / i;
  return res;
}
int main() {
  int a = read(), b = read(), c = read(), d = min(b, c);
  unsigned int ans = 0;
  mu[1] = 1;
  for (int i = (1); i <= (2000); i++)
    for (int j = i + i; j <= 2000; j += i) mu[j] -= mu[i];
  for (int i = (1); i <= (a); i++)
    for (int j = (1); j <= (d); j++)
      if (gcd(i, j) == 1) {
        ans += mu[j] * (a / i) * solve(i, b / j) * solve(i, c / j);
      }
  return printf("%u\n", ans % 1073741824), 0;
}