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;
unsigned niz[1000001] = {0};
int main() {
  unsigned i, j, k, n;
  unsigned d[101] = {0};
  unsigned rez = 0;
  const unsigned mod = 1073741824;
  set<int> s;
  for (i = 1; i < 101; i++)
    for (j = 1; j <= 100; j++)
      for (k = 1; k < 101; k++) niz[i * j * k] = 1;
  for (i = 1; i <= 1000000; i++)
    if (niz[i]) {
      k = 0;
      n = sqrt(double(i));
      for (j = 1; j <= n; j++)
        if (i % j == 0) k += 2;
      if (n * n == i) k--;
      niz[i] = k;
    }
  int a, b, c;
  cin >> a >> b >> c;
  for (i = 1; i <= a; i++)
    for (j = 1; j <= b; j++)
      for (k = 1; k <= c; k++) rez = (rez + niz[i * j * k]) % mod;
  cout << rez << endl;
  return 0;
}