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 long long mod = 1073741824;
int n, m, p, t = 0, pri[2010], g[2010][2010];
long long ans = 0, miu[2010];
bool vis[2010];
void shai(int n) {
  miu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) pri[++t] = i, miu[i] = -1;
    for (int j = 1; pri[j] * i <= n; j++) {
      vis[i * pri[j]] = 1;
      miu[i * pri[j]] = -miu[i];
      if (i % pri[j] == 0) {
        miu[i * pri[j]] = 0;
        break;
      }
    }
  }
}
long long gcd(int a, int b) {
  if (g[a][b]) return g[a][b];
  return g[a][b] = (b == 0 ? a : gcd(b, a % b));
}
long long f(int t, int k, int d) {
  long long ret = 0;
  for (int i = 1; i <= t / d; i++)
    if (gcd(max(d * i, k), min(d * i, k)) == 1)
      ret = (ret + 1ll * t / (d * i)) % mod;
  return ret;
}
int main() {
  shai(2000);
  scanf("%d%d%d", &n, &m, &p);
  if (n > m) swap(n, m);
  for (int k = 1; k <= p; k++)
    for (int d = 1; d <= n; d++)
      ans =
          (ans + 1ll * (p / k) * miu[d] * f(n, k, d) * f(m, k, d) + mod) % mod;
  printf("%lld", ans);
  return 0;
}