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;
const int INF = (int)1e9;
const double PI = 2 * acos(0.0);
const double eps = 1e-9;
const int NPOS = -1;
const int MAX = 1000002;
int a, b, c;
int64_t ans;
array<int, MAX> dev;
void DO() {
  for (int i = 1; i < MAX; ++i)
    for (int j = i; j < MAX; j += i) ++dev[j];
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  DO();
  cin >> a >> b >> c;
  for (int i = 1; i <= a; ++i)
    for (int j = 1; j <= b; ++j)
      for (int k = 1; k <= c; ++k) ans += dev[i * j * k], ans %= 1 << 30;
  cout << ans << '\n';
  return 0;
}