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;
vector<list<long long int> > m(1000005);
long long int prime[1000005];
void sieve() {
  prime[1] = 1;
  for (long int i = 2; i <= 1000003; i++) {
    if (prime[i] == 0) {
      for (long int j = 2; i * j <= 1000003; j++) {
        prime[i * j] = 1;
        long int temp = j;
        long int counter = 1;
        while (temp % i == 0) {
          counter++;
          temp = temp / i;
        }
        m[i * j].push_back(counter + 1);
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  long int a, b, c;
  cin >> a >> b >> c;
  sieve();
  long long int mod = (long long int)(1073741824 * (pow(2, 30)));
  for (long int i = 1; i <= 1000003; i++) {
    long int ans = 1;
    list<long long int>::iterator iter;
    if (prime[i] == 0) {
      prime[i] = 2;
      continue;
    }
    for (iter = m[i].begin(); iter != m[i].end(); iter++) {
      ans = (ans * (*iter)) % mod;
    }
    prime[i] = ans;
  }
  long long int sum = 0;
  for (long int i = 1; i <= a; i++) {
    for (long int j = 1; j <= b; j++) {
      for (long int k = 1; k <= c; k++) {
        sum += (prime[i * j * k]) % mod;
      }
    }
  }
  cout << sum << endl;
  return 0;
}