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 = 2000;
const int mod = 1 << 30;
int tot, p[N + 5], mu[N + 5], g[N + 5][N + 5];
bool flg[N + 5];
int gcd(int a, int b) {
  if (g[a][b]) return g[a][b];
  return g[a][b] = (b ? gcd(b, a % b) : a);
}
void init() {
  mu[1] = 1;
  for (int i = 2; i <= N; ++i) {
    if (!flg[i]) p[++tot] = i, mu[i] = -1;
    for (int j = 1; j <= tot && i * p[j] <= N; ++j) {
      flg[i * p[j]] = 1;
      if (i % p[j] == 0) {
        mu[i * p[j]] = 0;
        break;
      }
      mu[i * p[j]] = -mu[i];
    }
  }
}
long long calc(int d, int x) {
  long long res = 0;
  for (int i = 1; i <= d; ++i)
    if (gcd(i, x) == 1) res += d / i;
  return res;
}
int main() {
  init();
  int a, b, c;
  long long ans = 0;
  scanf("%d%d%d", &a, &b, &c);
  for (int i = 1; i <= a; ++i)
    for (int j = 1; j <= std::min(b, c); ++j)
      if (gcd(i, j) == 1)
        (ans += 1LL * (a / i) * mu[j] * calc(b / j, i) % mod * calc(c / j, i) %
                mod) %= mod;
  printf("%I64d\n", (ans % mod + mod) % mod);
  return 0;
}