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;
int primes[1000010];
int divs[1000010];
void pre() {
  memset(primes, -1, sizeof primes);
  for (int i = 2; i * i <= 1000010; i++) {
    if (primes[i] == -1)
      for (int j = 2; j * i < 1000010; j++) primes[j * i] = i;
  }
  memset(divs, -1, sizeof divs);
  divs[1] = 1;
}
int numDiv(int x) {
  if (divs[x] != -1) return divs[x];
  map<int, int> m;
  int p = x;
  while (primes[x] != -1) {
    m[primes[x]]++;
    x /= primes[x];
  }
  m[x]++;
  int resp = 1;
  for (typeof(m).begin() i = (m).begin(); i != (m).end(); i++)
    resp *= (i->second + 1);
  return divs[p] = resp;
}
int main() {
  pre();
  int a, b, c;
  while (scanf("%d%d%d", &a, &b, &c) == 3) {
    int cont = 0;
    for (int i = 1; i < a + 1; i++)
      for (int j = 1; j < b + 1; j++)
        for (int k = 1; k < c + 1; k++) {
          int aux = numDiv(i * j * k);
          cont = (cont % 1073741824 + aux % 1073741824) % 1073741824;
        }
    printf("%d\n", cont);
  }
  return 0;
}