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 N = (1 << 30) - 1;
int a, b, c;
int gcd[2005][2005];
inline bool sq_free(int n) {
  int j;
  for (int i = 2; (j = i * i) <= n; ++i)
    if (n % j == 0) return false;
  return true;
}
inline int sign(int n) {
  int cnt = 0;
  for (int i = 2; i * i <= n; ++i)
    if (n % i == 0) {
      cnt++;
      n /= i;
    }
  if (n > 1) cnt++;
  if (cnt & 1) return -1;
  return 1;
}
inline int calc(int x, int n) {
  int res = 0;
  for (int i = 1; i <= n; ++i)
    if (gcd[x][i] == 1) res += n / i;
  return res;
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  for (int i = 1; i <= 2000; ++i) gcd[i][0] = i;
  for (int i = 1; i <= 2000; ++i)
    for (int j = i; j <= 2000; ++j) gcd[i][j] = gcd[j][i] = gcd[i][j % i];
  int res = 0;
  for (int j = 1; j <= 2000; ++j)
    if (sq_free(j)) {
      int sq = sign(j);
      for (int i = 1; i <= a; ++i)
        if (gcd[i][j] == 1)
          res =
              (res + 1LL * sq * (a / i) * calc(i, b / j) * calc(i, c / j)) & N;
    }
  cout << res;
  return 0;
}