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>
using namespace std;
int a[1000010];
void solve() {
  int i, j;
  for (i = 1; i <= 1000000; i++) {
    if (i == 1)
      a[i] = 1;
    else
      a[i] = 2;
  }
  for (i = 2; i * i <= 1000000; i++) {
    for (j = i; j * i <= 1000000; j++) {
      if (i != j)
        a[i * j] += 2;
      else
        a[i * j]++;
    }
  }
}
int main() {
  int n, m, t, i, j, k, s;
  solve();
  while (~scanf("%d %d %d", &n, &m, &t)) {
    s = 0;
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) {
        for (k = 1; k <= t; k++) {
          s = (s + a[i * j * k]) % 1073741824;
        }
      }
    }
    printf("%d\n", s);
  }
  return 0;
}
