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;
bool primes[1000006];
unsigned long long divis[1000006];
void sieve() {
  for (int i = 1; i <= 1000006; ++i) primes[i] = 1;
  for (int i = 2; i * i <= 1000006; i++) {
    if (primes[i])
      for (int j = 2 * i; j <= 1000006; j += i) primes[j] = 0;
  }
}
void noDivis() {
  for (int i = 1; i < 1000006; ++i)
    for (int j = i; j < 1000006; j += i) divis[j]++;
}
int main() {
  int a, b, c;
  cin >> a >> b >> c;
  for (int i = 1; i < 1000006; ++i) divis[i] = 0;
  noDivis();
  unsigned long long s = 0;
  for (int i = 1; i <= a; ++i)
    for (int j = 1; j <= b; ++j)
      for (int k = 1; k <= c; ++k) {
        long temp = i * j * k;
        s += divis[temp];
        s %= 1073741824LL;
      }
  cout << s << endl;
  return 0;
}