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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int mod = 1073741824;
  char isprime[1000001];
  memset(isprime, 'y', sizeof(isprime));
  vector<int> primes;
  for (int i = 2; i < 1000001; i++) {
    if (isprime[i] == 'n') continue;
    int j = 2 * i;
    primes.push_back(i);
    while (j < 1000001) {
      isprime[j] = 'n';
      j += i;
    }
  }
  int ans = 0;
  int a, b, c;
  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++) {
        int p = i * j * k;
        if (p == 1) {
          ans += 1;
          if (ans >= mod) ans -= mod;
          continue;
        }
        if (isprime[p] == 'y') {
          ans += 2;
          if (ans >= mod) ans -= mod;
          continue;
        }
        int lmt = sqrt(p);
        int tmp = 1;
        for (int x = 0; primes[x] <= p && primes[x] <= lmt; x++) {
          int cnt = 0;
          while (p % primes[x] == 0) {
            p /= primes[x];
            cnt++;
          }
          tmp *= (cnt + 1);
        }
        if (p > 1) tmp *= 2;
        ans += tmp;
        if (ans > mod) ans -= mod;
      }
    }
  }
  cout << ans << "\n";
  return 0;
}
