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 ceill(long long a, long long b) { return a / b + bool(a % b); }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    x = (x * x) % 1000000007;
    y >>= 1;
  }
  return res;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
void primeFactors(long long n, vector<long long> &vec) {
  while (n % 2 == 0) vec.push_back(2), n /= 2;
  for (long long i = 3; i <= sqrt(n); i += 2)
    while (n % i == 0) vec.push_back(i), n /= i;
  if (n > 2) vec.push_back(n);
}
long long dp[1000005];
void pre() {
  for (long long i = 1; i <= 1000000; i++)
    for (long long j = i; j <= 1000000; j += i) dp[j]++;
}
void solve() {
  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 <= b; j++) {
      for (long long k = 1; k <= c; k++) {
        long long tmp = i * j * k;
        ans += (dp[tmp]) % 1073741824;
        ans %= 1073741824;
      }
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pre();
  long long tc = 1;
  for (long long TTT = 1; TTT <= tc; TTT++) {
    solve();
  }
}
