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 ≤ 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>
using namespace std;
int arr[10000000] = {0};
int main() {
  long long int a, b, c, l;
  scanf("%lld %lld %lld", &a, &b, &c);
  long long int i, j, k, sum = 0;
  for (i = 1; i <= a; i++) {
    for (j = 1; j <= b; j++) {
      for (k = 1; k <= c; k++) {
        long long int num = i * j * k;
        if (arr[num] != 0) {
          sum = (arr[num] % 1073741824) + (sum % 1073741824);
          continue;
        }
        long long int cnt = 0;
        long long int r = sqrt(num * 1.0);
        for (l = 1; l <= r; l++) {
          if (num % l == 0) {
            cnt++;
          }
        }
        if (r * r == num) {
          cnt = 2 * cnt - 1;
        } else {
          cnt = 2 * cnt;
        }
        arr[num] = cnt;
        sum = (cnt % 1073741824) + (sum % 1073741824);
      }
    }
  }
  printf("%lld\n", sum % 1073741824);
  return 0;
}
