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>
const int N = 2005;
int a, b, c, tot, p[N], mu[N], g[N][N];
bool np[N] = {true, true};
int gcd(int a, int b) {
  if (b == 0) return a;
  if (g[a][b]) return g[a][b];
  return g[a][b] = gcd(b, a % b);
}
int F(int n, int m) {
  int ret = 0;
  for (int i = 1; i <= n; i++)
    if (gcd(i, m) == 1) ret += n / i;
  return ret;
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  mu[1] = 1;
  for (int i = 2; i <= 2000; i++) {
    if (!np[i]) p[++tot] = i, mu[i] = -1;
    for (int j = 1; j <= 2000 && p[j] * i <= b; j++) {
      np[p[j] * i] = true;
      if (i % p[j] == 0) break;
      mu[i * p[j]] = -mu[i];
    }
  }
  int ans = 0;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b && j <= c; j++)
      if (gcd(i, j) == 1) ans += a / i * mu[j] * F(b / j, i) * F(c / j, i);
  printf("%d\n", ans & ((1 << 30) - 1));
}