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 ≤ 100).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Examples

Input

2 2 2


Output

20


Input

5 6 7


Output

1520

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.
#include <bits/stdc++.h>
using namespace std;
long long mod = 1ll << 30;
void __p(long long x) { cout << x; }
void __p(int32_t x) { cout << x; }
void __p(float x) { cout << x; }
void __p(double x) { cout << x; }
void __p(long double x) { cout << x; }
void __p(char x) { cout << '\'' << x << '\''; }
void __p(const char *x) { cout << '\"' << x << '\"'; }
void __p(const string &x) { cout << '\"' << x << '\"'; }
void __p(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __p(const pair<T, V> &x) {
  cout << '{';
  __p(x.first);
  cout << ',';
  __p(x.second);
  cout << '}';
}
template <typename T>
void __p(const T &x) {
  long long f = 0;
  cout << '{';
  for (auto &i : x) cout << (f++ ? "," : ""), __p(i);
  cout << "}";
}
void _p() { cout << "]\n"; }
template <typename T, typename... V>
void _p(T t, V... v) {
  __p(t);
  if (sizeof...(v)) cout << ", ";
  _p(v...);
}
long long po(long long x, long long p, long long m = 1e18) {
  long long ans = 1;
  x %= m;
  while (p > 0) {
    if (p & 1) ans = (ans * x) % m;
    p = p >> 1;
    x = (x * x) % m;
  }
  return ans;
}
long long inv(long long x, long long m = mod) { return po(x, m - 2, m) % m; }
const long long N = 1e6 + 6, M = N;
long long n, m, k, x, y, z, a1, b, c, d, i, j, l, r;
long long get(long long n) {
  long long x = 1;
  map<long long, long long> m;
  while (n % 2 == 0) {
    n /= 2;
    m[2]++;
  }
  for (long long i = 3; i * i <= n; i += 2) {
    while (n % i == 0) {
      n /= i;
      m[i]++;
    }
  }
  if (n > 1) m[n]++;
  for (auto p : m) {
    x *= (p.second + 1);
  }
  return x;
}
void sol() {
  long long a;
  cin >> a >> b >> c;
  x = 0;
  for (i = 1; i <= a; i++)
    for (j = 1; j <= b; j++)
      for (k = 1; k <= c; k++) {
        n = i * j * k;
        x = (x + get(n)) % mod;
      }
  cout << x << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  sol();
  return 0;
}
