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;
const int max_n = 2000 + 10;
const int max_n_2 = max_n * max_n;
const int mod = 1073741824;
bool check[max_n];
int prime[max_n], mu[max_n], tot;
int gcd[max_n][max_n];
long long _f[max_n], f[max_n_2];
inline void init(int m) {
  mu[1] = 1;
  for (int i = 2; i <= m; i++) {
    if (!check[i]) {
      prime[++tot] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= tot && prime[j] * i <= m; j++) {
      check[prime[j] * i] = true;
      if (i % prime[j] == 0) {
        mu[prime[j] * i] = 0;
        break;
      } else {
        mu[prime[j] * i] = -mu[i];
      }
    }
  }
}
inline int getnum() {
  int ans = 0;
  char c;
  bool flag = false;
  while (!isdigit(c = getchar()) && c != '-')
    ;
  if (c == '-')
    flag = true;
  else
    ans = c - '0';
  while (isdigit(c = getchar())) ans = ans * 10 + c - '0';
  return ans * (flag ? -1 : 1);
}
int get_gcd(int a, int b) {
  if (!b) return a;
  if (gcd[a][b]) return gcd[a][b];
  return gcd[a][b] = get_gcd(b, a % b);
}
int main() {
  init(2000);
  int a = getnum(), b = getnum(), c = getnum();
  int ab = a * b;
  for (int i = 1; i <= c; i++) {
    for (int j = i; j <= c; j += i) _f[i] += c / j;
    _f[i] *= mu[i];
  }
  for (int i = 1; i <= c; i++)
    if (mu[i])
      for (int j = i; j <= ab; j += i) f[j] += _f[i];
  long long ans = 0;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++)
      if (get_gcd(i, j) == 1) ans += (a / i) * (b / j) * f[i * j];
  cout << ans % mod << endl;
}