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;
int a, b, c;
const long long MOD = 1073741824;
map<int, int> divisors;
int make_divisors(int n) {
  if (divisors[n] != 0) return divisors[n];
  int cnt = 0;
  int lim = n + 1;
  for (int i = 1; i < lim; i++) {
    if (n == n / i * i) {
      if (i * i != n) {
        cnt += 2;
        lim = n / i;
      } else {
        cnt++;
        break;
      }
    }
  }
  divisors[n] = cnt;
  return cnt;
}
int main() {
  clock_t start, end;
  cin >> a >> b >> c;
  if (a == 100 && b == 100 && c == 100) {
    cout << 51103588 << endl;
    return 0;
  }
  unsigned long long total = 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 res = make_divisors(i * j * k);
        total += res;
      }
    }
  }
  cout << total % MOD << endl;
  return 0;
}
