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;
bool iss(int d) {
  for (int i = 2; i * i <= d; ++i)
    if (d % i == 0) return false;
  return true;
}
int main() {
  int a, b, c;
  scanf("%d%d%d", &a, &b, &c);
  vector<int> simple;
  for (int i = 2; i < 100; ++i)
    if (iss(i)) simple.push_back(i);
  vector<vector<int> > d(101);
  for (int i = 1; i <= 100; ++i) {
    d[i].resize(simple.size());
    for (int j = 0, tmp = i; j < simple.size(); ++j)
      while (tmp % simple[j] == 0) {
        tmp /= simple[j];
        d[i][j]++;
      }
  }
  int ans = 0;
  int mod = 1 << 30;
  for (int i = 1; i <= a; ++i)
    for (int j = 1; j <= b; ++j)
      for (int k = 1; k <= c; ++k) {
        int prod = 1;
        for (int l = 0; l < simple.size(); ++l)
          prod *= d[i][l] + d[j][l] + d[k][l] + 1;
        ans = (ans + prod) % mod;
      }
  printf("%d\n", ans);
  return 0;
}
