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:

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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 ≤ 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;
long long dp[(int)1e6 + 5];
long long factorize(long long n) {
  if (dp[n] != -1) return dp[n];
  long long ans = 0;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      if (i * i != n)
        ans += 2;
      else
        ans++;
    }
  }
  dp[n] = ans;
  return ans;
}
int main() {
  long long a, b, c;
  cin >> a >> b >> c;
  memset(dp, -1, sizeof dp);
  long long m = 1 << 30;
  long long ans = 0;
  for (long long i = 1; i <= a; i++) {
    for (long long j = 1; j <= b; j++) {
      for (long long k = 1; k <= c; k++) {
        long long product = i * j * k;
        ans = (ans % m + (int)factorize(product) % m) % m;
      }
    }
  }
  cout << ans;
  return 0;
}