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:

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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 ≤ 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 mod = 1 << 30;
const int C = 3000;
bool prime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return x > 1;
}
int get_div(int x) {
  int ans = 0;
  for (int i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      ans += 1 + (i * i < x);
    }
  }
  return ans;
}
unordered_map<long long, unsigned int> d, tmp;
int main() {
  d.rehash(77777);
  tmp.rehash(77777);
  int a, b, c;
  cin >> a >> b >> c;
  if (a > b) swap(a, b);
  if (a > c) swap(a, c);
  if (b > c) swap(b, c);
  d[(a * C + b) * 1ll * C + c] = 1;
  for (int i = 2000; i > 1; i--) {
    if (prime(i)) {
      tmp.clear();
      for (auto &it : d) {
        int a = (it.first / C) / C;
        int b = (it.first / C) % C;
        int c = it.first % C;
        for (int dega = 0, pa = a; pa; pa /= i, dega++) {
          for (int degb = 0, push_back = b; push_back; push_back /= i, degb++) {
            long long tt = (pa * C + push_back) * 1ll * C;
            for (int degc = 0, pc = c; pc; pc /= i, degc++) {
              tmp[tt + pc] += it.second * (dega + degb + degc + 1);
            }
          }
        }
      }
      d.clear();
      for (auto it : tmp) {
        int a = (it.first / C) / C;
        int b = (it.first / C) % C;
        int c = it.first % C;
        if (a > b) swap(a, b);
        if (a > c) swap(a, c);
        if (b > c) swap(b, c);
        d[(a * C + b) * 1ll * C + c] += it.second % mod;
      }
    }
  }
  long long ans = 0;
  for (auto &it : d) {
    ans += it.second;
  }
  cout << ans % mod;
  return 0;
}