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;
vector<int> v;
int ma = 0;
bool p[10000000];
vector<int> primes;
void Prime() {
  memset(p, -1, sizeof p);
  p[0] = 0;
  p[1] = 1;
  for (int i = 2; i < 10000000; i++) {
    if (p[i]) {
      primes.push_back(i);
      for (int j = i * 2; j < 10000000; j += i) {
        p[j] = 0;
      }
    }
  }
}
int Divisors2(int n) {
  int d = 1, e = 0;
  for (int p : primes) {
    if (p * p > n) break;
    while (n % p == 0) {
      e++;
      n /= p;
    }
    d *= (e + 1);
    e = 0;
  }
  if (n != 1) d *= 2;
  return d;
}
int main() {
  Prime();
  int a, b, c;
  long long sum = 0;
  cin >> a >> b >> c;
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        v.push_back(i * j * k);
      }
    }
  }
  for (int i = 0; i < (int)v.size(); i++) {
    sum += Divisors2(v[i]);
  }
  cout << sum;
  return 0;
}