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;
unsigned int n, a, b, c, f[4010101] = {0}, ans = 0;
int pn = 0, p[2010], mu[2010];
bool isp[2010];
void init() {
  memset(isp, true, sizeof isp);
  isp[1] = false;
  mu[1] = 1;
  for (int i = 2; i <= a; ++i) {
    if (isp[i]) p[pn++] = i, mu[i] = -1;
    for (int j = 0; i * p[j] <= a; ++j) {
      isp[i * p[j]] = false;
      if (i % p[j] == 0) {
        mu[i * p[j]] = 0;
        break;
      }
      mu[i * p[j]] = -mu[i];
    }
  }
}
int main() {
  cin >> a >> b >> c;
  init();
  n = b * c;
  for (int i = 1; i <= b; ++i)
    for (int j = 1; j <= c; ++j) ++f[i * j];
  for (int k = 0; k < 2; ++k)
    for (int i = 1; i <= n; ++i)
      for (int j = i + i; j <= n; j += i) f[i] += f[j];
  for (int i = 1; i <= a; ++i)
    for (int j = i; j <= a; j += i) ans += mu[i] * f[i] * (a / j);
  cout << ans % 1073741824 << endl;
  return 0;
}