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;
const int MOD = (1 << 30) - 1, MAX = 1e6 + 2;
int divisors[MAX];
int getDivisors(int n) {
  int cnt = 0;
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ++cnt;
      if (i * i != n) ++cnt;
    }
  }
  return cnt;
}
int main() {
  int a, b, c;
  long long sum = 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++) {
        long long tmp = 1ll * i * j * k;
        if (divisors[tmp] == 0) divisors[tmp] = getDivisors(tmp);
        sum += divisors[tmp];
      }
  cout << sum % MOD;
}
