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;
struct node {
  long long f[26];
} g[111];
long long a[30], tot;
bool b[110];
void getprim() {
  b[1] = true;
  for (long long i = 2; i <= 101; i++) {
    if (b[i]) continue;
    a[++tot] = i;
    for (long long j = 2 * i; j <= 101; j += i) b[j] = true;
  }
}
int main() {
  long long total, alltotal;
  long long k, AA, BB, CC;
  long long tep;
  getprim();
  for (long long i = 1; i <= 100; i++) {
    k = i;
    memset(g[i].f, 0, sizeof g[i].f);
    for (long long j = 1; j <= tot; j++)
      while (k % a[j] == 0) g[i].f[j]++, k /= a[j];
  }
  alltotal = 0;
  cin >> AA >> BB >> CC;
  for (long long A = 1; A <= AA; A++)
    for (long long B = 1; B <= BB; B++)
      for (long long C = 1; C <= CC; C++) {
        total = 1;
        for (long long i = 1; i < tot; i++) {
          tep = g[A].f[i] + g[B].f[i] + g[C].f[i];
          if (tep > 0) total = (total * (tep + 1)) % 1073741824;
        }
        alltotal = (alltotal + total) % 1073741824;
      }
  cout << alltotal << endl;
  return 0;
}
