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;
int pri[5005], pcnt, mu[5005];
bool notp[5005];
int gcd[5005][5005];
void Init(int n) {
  mu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!notp[i]) pri[++pcnt] = i, mu[i] = -1;
    for (int j = 1, t; j <= pcnt; j++) {
      t = i * pri[j];
      if (t > n) break;
      notp[t] = 1;
      if (i % pri[j] == 0) {
        mu[t] = 0;
        break;
      }
      mu[t] = -mu[i];
    }
  }
  for (int i = 1; i <= n; i++) {
    gcd[i][i] = gcd[i][0] = gcd[0][i] = i;
    for (int j = 1; j < i; j++) gcd[j][i] = gcd[i][j] = gcd[j][i % j];
  }
}
unsigned int work(int n, int d, int k) {
  unsigned int res = 0;
  for (int i = 1; i <= n / d; i++)
    if (gcd[i * d][k] == 1) res += n / (i * d);
  return res;
}
int main() {
  int A, B, C;
  cin >> A >> B >> C;
  int n = max(max(A, B), C);
  Init(n);
  unsigned int res = 0;
  for (int k = 1; k <= C; k++) {
    unsigned int tmp = 0;
    for (int d = 1; d <= n; d++)
      if (mu[d]) tmp += (unsigned int)mu[d] * work(A, d, k) * work(B, d, k);
    res += tmp * (C / k);
  }
  cout << (res % (1 << 30)) << endl;
  return 0;
}