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;
unsigned long long arr[1000005];
unsigned long long divCount(int n) {
  if (n == 1) return 1;
  unsigned long long count = 1;
  int i;
  for (i = 2; i * i < n; ++i) {
    if (n % i == 0) ++count;
  }
  count *= 2;
  if (i * i == n) {
    ++count;
  }
  return count;
}
int main() {
  int a, b, c;
  unsigned long long sum = 0;
  cin >> a >> b >> c;
  for (int i = a; i > 0; i--) {
    for (int j = b; j > 0; j--) {
      for (int k = c; k > 0; k--) {
        if (arr[i * j * k] == 0) {
          arr[i * j * k] = divCount(i * j * k);
        }
        sum = sum + (arr[i * j * k] % 1073741824);
      }
    }
  }
  cout << sum << " \n";
  return 0;
}