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 IN() {
  char c;
  register int first = 0;
  while ((c = getchar()) < 48 && c ^ '-' || c > 57)
    ;
  bool f = c == '-';
  if (f) (c = getchar());
  for (; c > 47 && c < 58; (c = getchar())) first = first * 10 + c - 48;
  if (f) first = -first;
  return first;
}
int A, B, C;
int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
inline bool sqf(int first) {
  for (int i = 2; i * i <= first; ++i)
    if (first % (i * i) == 0) return 0;
  return 1;
}
inline int get_mu(int first) {
  int re = 1;
  for (int i = 2; i * i <= first; ++i)
    if (first % i == 0) re *= -1, first /= i;
  return first > 1 ? -re : re;
}
inline int suan(int _, int N) {
  int re = 0;
  for (int _r = N, i = 1; i <= _r; ++i)
    if (gcd(_, i) == 1) re += N / i;
  return re;
}
int main() {
  A = IN();
  B = IN();
  C = IN();
  long long an = 0;
  for (int _r = A, d = 1; d <= _r; ++d)
    if (sqf(d)) {
      int mu = get_mu(d);
      for (int _r = C, k = 1; k <= _r; ++k)
        if (gcd(d, k) == 1)
          an += (long long)(C / k) * (long long)mu * (long long)suan(k, A / d) *
                (long long)suan(k, B / d);
    }
  printf("%d\n", an % (1 << 30));
}