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() {
  int a, b, c;
  cin >> a >> b >> c;
  int v[101][25];
  int arr[25] = {2,  3,  5,  7,  11, 13, 17, 19, 23, 29, 31, 37, 41,
                 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};
  for (int i = 1; i < 101; i++) {
    for (int j = 0; j < 25; j++) {
      int ans = 0;
      int temp = i;
      while ((temp % arr[j]) == 0) {
        temp = temp / arr[j];
        ans++;
      }
      v[i][j] = ans;
    }
  }
  long long int ans = 0;
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        long long int temp = 1;
        for (int l = 0; l < 25; l++) {
          temp *= (1 + v[i][l] + v[j][l] + v[k][l]);
        }
        ans = (ans + temp) % 1073741824;
      }
    }
  }
  cout << ans;
}
