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:

<image>

Find the sum modulo 1073741824 (230).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 100).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Examples

Input

2 2 2


Output

20


Input

5 6 7


Output

1520

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.
#include <bits/stdc++.h>
int arr[1000005];
using namespace std;
int main() {
  int i, j, k, number, sn, n, m, l;
  long long sum;
  for (i = 2; i * i < 1000005; i++) {
    for (j = i * i + i; j < 1000005; j += i) {
      arr[j]++;
    }
  }
  scanf("%d", &m);
  scanf("%d", &n);
  scanf("%d", &l);
  sum = 0;
  for (i = 1; i <= m; i++) {
    for (j = 1; j <= n; j++) {
      for (k = 1; k <= l; k++) {
        number = i * j * k;
        sn = (int)sqrt(number);
        if (sn * sn == number) {
          sum = (sum + 2 * (arr[number]) + 1 + 2) % 1073741824;
        } else {
          sum = (sum + 2 * arr[number] + 2) % 1073741824;
        }
      }
    }
  }
  sum = (sum - 2) % 1073741824;
  cout << sum;
}
