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 vector<int> primes = {2,  3,  5,  7,  11, 13, 17, 19, 23, 29, 31, 37, 41,
                            43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};
vector<unordered_map<int, int>> powers(101);
void calc() {
  for (int i = 2; i <= 100; i++) {
    for (int j = i, p = 0; j > 1 && p < primes.size(); p++) {
      int c = 0;
      for (; j % primes[p] == 0; j /= primes[p]) c++;
      if (c) powers[i][primes[p]] = c;
    }
  }
}
int divisors(int i, int j, int k) {
  unordered_map<int, int> r;
  for (auto [k, v] : powers[i]) r[k] += v;
  for (auto [k, v] : powers[j]) r[k] += v;
  for (auto [k, v] : powers[k]) r[k] += v;
  int d = 1;
  for (auto [k, v] : r) d *= v + 1;
  return d;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  calc();
  int a, b, c;
  cin >> a >> b >> c;
  int s = 0;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++)
      for (int k = 1; k <= c; k++) {
        s += divisors(i, j, k);
        if (s > 1073741824) s -= 1073741824;
      }
  cout << s << '\n';
}
