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>
int arr[1000001];
int main() {
  int a, b, c, ans = 0;
  std::cin >> a >> b >> c;
  for (int i = 1; i <= a * b * c; ++i) {
    for (int j = i; j <= a * b * c; j += i) {
      ++arr[j];
    }
  }
  for (int i = 1; i <= a; ++i) {
    for (int j = 1; j <= b; ++j) {
      for (int k = 1; k <= c; ++k) {
        ans = (ans + arr[i * j * k]) % (1 << 30);
      }
    }
  }
  std::cout << ans << std::endl;
  return 0;
}