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 N = 500005;
const long long modd = 1073741824;
long long check[N + 5];
long long mu[N + 5], p[N + 5];
bool flg[N + 5];
long long g[2002][2002];
long long gcd(long long a, long long b) {
  if (g[a][b])
    return g[a][b];
  else
    return g[a][b] = g[b][a] = (b ? gcd(b, a % b) : a);
}
void mobius() {
  long long tot = 0;
  mu[1] = 1;
  for (long long i = 2; i <= N; ++i) {
    if (!flg[i]) {
      p[++tot] = i;
      mu[i] = -1;
    }
    for (long long 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 solve(long long a, long long b, long long c) {
  if (!g[a][b]) gcd(a, b);
  if (g[a][b] != 1) return 0;
  long long res = 0;
  for (long long i = 1; i <= c / a; i++) {
    if (!g[i][b]) gcd(i, b);
    if (g[i][b] != 1) continue;
    res += c / (i * a);
    res %= modd;
  }
  return res;
}
int main() {
  mobius();
  long long a, b, c;
  cin >> a >> b >> c;
  long long ans = 0;
  for (long long i = 1; i <= a; i++) {
    for (long long j = 1; j <= min(b, c); j++) {
      if (g[i][j] == 0) gcd(i, j);
      if (g[i][j] == 1) {
        ans += (a / i % modd) * (mu[j]) *
               ((solve(j, i, b) % modd) * (solve(j, i, c) % modd)) % modd;
        ans %= modd;
      }
    }
  }
  cout << ans << endl;
}