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 mod = 1073741824;
int a, b, c, num;
int prime[2005], miu[2005];
bool check[2005];
long long ans;
inline int gcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a > b)
    return gcd(a % b, b);
  else
    return gcd(b % a, a);
}
inline void mobius() {
  miu[1] = 1;
  for (int i = 2; i <= max(a, max(b, c)); i++) {
    if (!check[i]) miu[i] = -1, prime[++num] = i;
    for (int j = 1; j <= num && prime[j] * i <= max(a, max(b, c)); j++) {
      check[prime[j] * i] = 1;
      miu[prime[j] * i] = -miu[i];
      if (i % prime[j] == 0) {
        miu[prime[j] * i] = 0;
        break;
      }
    }
  }
}
inline int f(int d, int x) {
  int ans = 0;
  for (int i = 1; i <= d; i++)
    if (gcd(i, x) == 1) ans += d / i;
  return ans;
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  mobius();
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++)
      if (gcd(i, j) == 1)
        ans = (ans + (long long)miu[j] * (a / i) * f(b / j, i) * f(c / j, i)) %
              mod;
  printf("%I64d\n", ans);
  return 0;
}