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:

<image>

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;
const int N = 200 * 1000 + 15;
char mat[4][4];
int a, b, c;
int memo[1000 * 4 + 1];
int solve(int n) {
  if (n < 0) return -1e5;
  if (n == 0) return 0;
  if (memo[n] != -1) return memo[n];
  return memo[n] =
             max(solve(n - a) + 1, max(solve(n - b) + 1, solve(n - c) + 1));
}
long long _sieve_size = 1e6 + 2;
;
bitset<10000010> bs;
vector<int> primes;
int divisores[1000000 + 3];
void sieve() {
  bs[0] = bs[1] = 0;
  for (long long i = 2; i <= _sieve_size; i++) {
    for (long long j = i; j <= _sieve_size; j += i) {
      divisores[j]++;
    }
  }
}
int main() {
  int a, b, c;
  cin >> a >> b >> c;
  fill_n(divisores, _sieve_size, 1);
  sieve();
  long long total = 0;
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        total += divisores[i * j * k];
      }
    }
  }
  cout << total;
}