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;
int d(int x) {
  int T = 1;
  if (x == 1)
    return 1;
  else {
    for (int i = 1; i < x; i++) {
      if (x % i == 0) T++;
    }
    return T;
  }
}
int d1(int x) {
  int T = 0;
  for (int i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      if (i != (x / i)) T++;
      T++;
    }
  }
  return T;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t, n, m, x, y, k, L, T;
  vector<int> v1;
  set<int> set1;
  map<int, int> m1;
  map<int, bool> m2;
  map<string, int> mIndex;
  string s, ss, ss1;
  cin >> x >> y >> k;
  int TT = 0;
  for (int i = 1; i <= x; i++)
    for (int ii = 1; ii <= y; ii++)
      for (int iii = 1; iii <= k; iii++) {
        int k = (i * ii * (iii)*1LL);
        if (!m1[k]) m1[k] = d1(k);
        TT = (TT + m1[k]) % 1073741824;
      }
  cout << TT;
  return 0;
}
