Problem: 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 ≤ 2000).

Output

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

Examples

Input

2 2 2


Output

20


Input

4 4 4


Output

328


Input

10 10 10


Output

11536

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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
int t[1000000];
int d(int x) {
  if (t[x] != 0) return t[x];
  int s = 0;
  while ((s + 1) * (s + 1) <= x) s++;
  int w = 0;
  for (int i = 1; i <= s; i++)
    if (x % i == 0) w += 2;
  if (s * s == x) w--;
  t[x] = w;
  return t[x];
}
int main() {
  int a, b, c;
  int M = 1 << 30;
  int w = 0;
  scanf("%d %d %d", &a, &b, &c);
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++)
      for (int k = 1; k <= c; k++) {
        w = (w + d(i * j * k)) % M;
      }
  printf("%d\n", w);
  return 0;
}