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;
const long long MOD = 1 << 30;
long long add(long long a, long long b) { return (a + b) % MOD; }
const int MAXN = 1100000;
int dd[MAXN];
int d(int k) { return dd[k]; }
void prec() {
  for (int a = 1; a < MAXN; ++a) {
    for (int b = a; b < MAXN; b += a) {
      dd[b]++;
    }
  }
}
int main() {
  prec();
  int A, B, C;
  long long sum = 0;
  cin >> A >> B >> C;
  for (int i = 1; i <= A; ++i) {
    for (int j = 1; j <= B; ++j) {
      for (int k = 1; k <= C; ++k) {
        sum = add(sum, d(i * j * k));
      }
    }
  }
  cout << sum << endl;
  return 0;
}
