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;
unsigned long long binpow(unsigned long long a, unsigned long long n) {
  if (n == 0) return 1;
  if (n % 2 == 1) {
    unsigned long long cur_1 = binpow(a, n - 1);
    unsigned long long cur_2 =
        ((cur_1 % 1073741824) * (a % 1073741824)) % 1073741824;
    return cur_2;
  } else {
    unsigned long long b = binpow(a, n / 2);
    unsigned long long cur_3 =
        ((b % 1073741824) * (b % 1073741824)) % 1073741824;
    return cur_3;
  }
}
int main() {
  long result[1000000] = {0};
  long i, j, p, t, n;
  for (i = 0; i < 1000000; i++) {
    p = i + 1;
    for (j = p; j <= 1000000; j += p) {
      result[j - 1]++;
    }
  }
  int a, b, c;
  cin >> a >> b >> c;
  unsigned long long count = 0;
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        count += result[i * j * k - 1];
        count %= 1073741824;
      }
    }
  }
  cout << count;
  return 0;
}
