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>
void fileio_in_out() {
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
}
void fileio_txt() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
int n, ans = 0, a, b, c;
int i, j, k, l;
int f[1111111];
int calc(int n) {
  int i;
  if (!f[n])
    for (i = 1; i <= sqrt(n); i++)
      if (n % i == 0) f[n] += (n == i * i) ? 1 : 2;
  return f[n];
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  memset(f, 0, sizeof(f));
  for (i = 1; i <= a; i++)
    for (j = 1; j <= b; j++)
      for (k = 1; k <= c; k++) ans = (ans + calc(i * j * k)) % 1073741824;
  printf("%d\n", ans);
  return 0;
}
