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 ≤ 100).

Output

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

Examples

Input

2 2 2


Output

20


Input

5 6 7


Output

1520

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.
#include <bits/stdc++.h>
using namespace std;
vector<int> primes;
map<int, int> dv;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  primes.push_back(2);
  for (int i = 3; i <= 100; i++) {
    int d = 0;
    for (int j = 2; j <= sqrt(i); j++) {
      if (i % j == 0) {
        d++;
        j = i;
      }
    }
    if (d == 0) primes.push_back(i);
  }
  for (int i = 2; i <= 1000000; i++) {
    int d = i;
    map<int, int> t;
    for (int j = 0; j < primes.size(); j++) {
      if (d % primes[j] == 0) {
        d /= primes[j];
        t[primes[j]]++;
        j = -1;
        if (d <= 1) j = primes.size();
      }
    }
    map<int, int>::iterator it = t.begin();
    int nd = 1;
    while (it != t.end()) {
      int x = it->second;
      x++;
      nd *= x;
      it++;
    }
    dv[i] = nd;
  }
  int a, b, c;
  cin >> a >> b >> c;
  long long ans = 1;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++)
      for (int k = 1; k <= c; k++) {
        ans += dv[(i * j * k)];
        ans %= 1073741824;
      }
  cout << ans;
  return 0;
}
