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:

<image>

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;
const long long N = 2005;
long long gcd[N][N];
long long mu[N];
long long pri[N], tot = 0;
bool ok[N];
void prepare() {
  memset(ok, true, sizeof(ok));
  mu[1] = 1;
  for (long long u = 2; u <= 2000; u++) {
    if (ok[u]) {
      pri[++tot] = u;
      mu[u] = -1;
    }
    for (long long i = 1; i <= tot; i++) {
      long long j = pri[i];
      if (u * j > 2000) break;
      ok[u * j] = false;
      if (u % j == 0) {
        mu[u * j] = 0;
        break;
      } else
        mu[u * j] = -mu[u];
    }
  }
  for (long long u = 0; u <= 2000; u++) gcd[u][0] = u, gcd[0][u] = u;
  for (long long u = 1; u <= 2000; u++)
    for (long long i = 1; i <= u; i++) gcd[i][u] = gcd[u][i] = gcd[i][u % i];
}
long long a, b, c;
long long solve(long long x, long long y, long long z) {
  long long lalal = 0;
  for (long long u = 1; u <= x; u++)
    if (gcd[z][u * y] == 1) lalal = lalal + x / u;
  return lalal;
}
int main() {
  prepare();
  scanf("%lld%lld%lld", &a, &b, &c);
  long long ans = 0;
  for (long long u = 1; u <= a; u++) {
    long long lalal = 0;
    for (long long i = 1; i <= min(b, c); i++) {
      lalal =
          lalal + mu[i] * solve(b / i, i, u) % MOD * solve(c / i, i, u) % MOD;
      lalal %= MOD;
    }
    ans = ans + (a / u) * lalal % MOD;
    ans = ans % MOD;
  }
  ans = (ans + MOD) % MOD;
  printf("%lld\n", ans);
  return 0;
}