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;
bool factor[101] = {false};
int numDiv(int a) {
  if (a == 1) return 1;
  int i = 2;
  int cntr = 0;
  for (; i * i < a; i++) {
    if (a % i == 0) {
      cntr += 2;
    }
  }
  if (i * i == a) cntr++;
  cntr += 2;
  return cntr;
}
int main() {
  int a, b, c;
  int mod = 2 << 29;
  cin >> a >> b >> c;
  int d[1000001] = {0};
  for (int i = 1; i <= a * b * c; i++) {
    for (int j = 1; i * j <= a * b * c; j++) d[i * j]++;
  }
  long long sm = 0;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++)
      for (int k = 1; k <= c; k++) sm += d[i * j * k];
  cout << sm << endl;
}