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;
long long d[1111111];
int main() {
  long long t, z = 0, l, i = 0, len;
  long long n, k = 0, m = 0, bam = 0, dan = 0, ans = 0;
  long long md = 1073741824, a, b, c, sum = 0;
  scanf("%lld %lld %lld", &a, &b, &c);
  for (l = 1; l < (a * b * c) + 1; l++)
    for (i = l; i <= a * b * c; i += l) d[i]++;
  for (l = 1; l < a + 1; l++)
    for (i = 1; i < b + 1; i++)
      for (k = 1; k < c + 1; k++) sum += (d[l * i * k] % md);
  printf("%lld", sum);
  return 0;
}
