Problem: 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:

<image>

Find the sum modulo 1073741824 (230).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 2000).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Examples

Input

2 2 2


Output

20


Input

4 4 4


Output

328


Input

10 10 10


Output

11536

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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int const N = 1e6 + 5, M = 2e3 + 5, OO = 0x3f3f3f3f, MOD = 1073741824;
int a, b, c, ans;
int arr[N];
int d(int n) {
  if (arr[n] != 0) return arr[n];
  int temp = 0;
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      (i * i == n) ? temp++ : temp += 2;
    }
  }
  return arr[n] = temp;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> a >> b >> c;
  memset(arr, 0, sizeof arr);
  for (int i = 1; i < N; i++) {
    for (int j = i; j < N; j += i) {
      arr[j]++;
    }
  }
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        ans += arr[i * j * k];
        ans %= MOD;
      }
    }
  }
  cout << ans << endl;
  return 0;
}