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;
void mohammed() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
}
int d(int x) {
  int c = 0;
  for (int i = 1; i <= sqrt(x); i++) {
    if (x % i == 0) {
      if (i * i == x) {
        c++;
      } else
        c += 2;
    }
  }
  return c;
}
int main() {
  mohammed();
  int a, b, c, s = 0;
  int arr[1000001] = {0};
  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 x = i * j * k;
        if (arr[x] == 0) {
          int y = d(x);
          arr[x] = y;
          s += y;
        } else
          s += arr[x];
        s %= 1073741824;
      }
    }
  }
  cout << s % 1073741824;
  return 0;
}
