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;
bool p[1000001];
vector<int> v;
void sieve() {
  memset(p, true, sizeof(p));
  for (int i = 2; i <= sqrt(1000001); i++) {
    if (p[i] == true) {
      for (int j = i * 2; j <= 1000001; j += i) p[j] = false;
    }
  }
  for (int i = 2; i <= 1000000; i++) {
    if (p[i] == true) v.push_back(i);
  }
}
int main() {
  sieve();
  int a, b, c;
  cin >> a >> b >> c;
  long long int ans = 0;
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        int sum = (i * j * k);
        if (sum == 1)
          ans += 1;
        else if (p[sum] == true)
          ans += 2;
        else {
          long long int mul = 1;
          for (int l = 0; v[l] <= sum; l++) {
            int count = 0;
            while (sum % v[l] == 0) {
              sum /= v[l];
              count++;
              if (sum == 1) break;
            }
            if (count != 0) mul *= ((count + 1) % 1073741824);
          }
          ans += (mul) % 1073741824;
        }
      }
    }
  }
  cout << ans << endl;
}