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;
unordered_map<int, int> first[101];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 1; i <= 100; i++) {
    int tmp = i;
    for (int j = 2; j * j <= tmp; j++)
      while (tmp % j == 0) tmp /= j, first[i][j]++;
    if (tmp != 1) first[i][tmp]++;
  }
  int a, b, c;
  scanf("%d%d%d", &a, &b, &c);
  unordered_map<int, int> tf;
  long long ans = 0;
  for (int i = 1; i <= a; i++) {
    for (pair<int, int> aa : first[i]) tf[aa.first] += aa.second;
    for (int j = 1; j <= b; j++) {
      for (pair<int, int> aa : first[j]) tf[aa.first] += aa.second;
      for (int k = 1; k <= c; k++) {
        for (pair<int, int> aa : first[k]) tf[aa.first] += aa.second;
        long long tot = 1;
        for (pair<int, int> aa : tf) tot *= aa.second + 1, tot %= 1073741824;
        ans += tot;
        ans %= 1073741824;
        for (pair<int, int> aa : first[k]) tf[aa.first] -= aa.second;
      }
      for (pair<int, int> aa : first[j]) tf[aa.first] -= aa.second;
    }
    for (pair<int, int> aa : first[i]) tf[aa.first] -= aa.second;
  }
  printf("%lld\n", ans);
}