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;
long long res;
const long long mod = 1073741824;
long long d(int a, int b, int c) {
  vector<int> q(102, 0);
  for (int i = 2; i <= a; i++)
    if (a % i == 0) {
      while (a % i == 0) {
        a /= i;
        q[i]++;
      }
    }
  for (int i = 2; i <= b; i++)
    if (b % i == 0) {
      while (b % i == 0) {
        b /= i;
        q[i]++;
      }
    }
  for (int i = 2; i <= c; i++)
    if (c % i == 0) {
      while (c % i == 0) {
        c /= i;
        q[i]++;
      }
    }
  long long res = 1;
  for (int i = 0; i < 102; i++) res = (res * (q[i] + 1)) % mod;
  return res;
}
int main() {
  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++) res = (res % mod + d(i, j, k) % mod) % mod;
  cout << res;
}
