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:

<image>

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 N = 2000 + 5;
const int mod = 1 << 30;
short gcd[N][N], mobius[N];
int ans, tot, a, b, c, t;
inline int calc(int x, short *p) {
  int rt = 0;
  for (int i = 1; i <= x; ++i)
    if (p[i] == 1) rt += x / i;
  return rt;
}
int main(int argc, char **argv) {
  for (int i = 1; i < N; ++i) gcd[0][i] = gcd[i][0] = i;
  for (int i = 1; i < N; ++i)
    for (int j = i; j < N; ++j) gcd[i][j] = gcd[j][i] = gcd[j % i][i];
  mobius[1] = 1;
  for (int i = 2; i < N; ++i)
    for (int j = 1; j < i; ++j)
      if (!(i % j)) mobius[i] -= mobius[j];
  scanf("%d%d%d", &a, &b, &c);
  t = min(b, c);
  for (int i = 1; i <= a; ++i) {
    tot = 0;
    for (int j = 1; j <= t; ++j)
      if (gcd[i][j] == 1 && mobius[j])
        tot += mobius[j] * calc(b / j, gcd[i]) * calc(c / j, gcd[i]);
    ans += tot * (a / i);
  }
  printf("%d\n", (ans % mod + mod) % mod);
  return 0;
}