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 M = 2e3 + 10;
const int mod = (1 << 30);
int a, b, c, n;
int p[M], pn = 0;
bool isnp[M];
int mu[M];
int GCD[M][M];
int gcd(int a, int b) {
  if (b == 0) return a;
  if (GCD[a][b])
    return GCD[a][b];
  else
    return GCD[a][b] = GCD[b][a] = gcd(b, a % b);
}
long long getans(int to, int d, int k) {
  long long res = 0;
  for (int i = 1; i <= to; ++i)
    if (gcd(d * i, k) == 1) res = res + int(to / i);
  return res % mod;
}
int main() {
  cin >> a >> b >> c;
  if (a > b) swap(a, b);
  if (a > c) swap(a, c);
  if (b > c) swap(b, c);
  n = c;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) gcd(i, j);
  mu[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!isnp[i]) {
      p[++pn] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= pn && i * p[j] <= n; ++j) {
      isnp[i * p[j]] = 1;
      if (i % p[j] == 0) {
        mu[i * p[j]] = 0;
        break;
      } else
        mu[i * p[j]] = -mu[i];
    }
  }
  long long ans = 0;
  for (int k = 1; k <= c; ++k) {
    long long res = 0;
    for (int d = 1; d <= a; ++d) {
      if (mu[d] == 0) continue;
      res += getans(a / d, d, k) * getans(b / d, d, k) % mod * mu[d];
    }
    res = res * int(c / k);
    ans += res % mod;
  }
  ans = (ans % mod + mod) % mod;
  cout << ans << endl;
  return 0;
}