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>
int a, b, c, prad[1000001];
using namespace std;
int main() {
  long long pro = 0;
  for (int i = 1; i < 1000001; ++i) {
    for (int j = 1; j * i < 1000001; ++j) {
      ++prad[j * i];
    }
  }
  cin >> a >> b >> c;
  for (int j = 1; j < a + 1; ++j)
    for (int ja = 1; ja < b + 1; ++ja)
      for (int ha = 1; ha < c + 1; ++ha) {
        pro += prad[ha * ja * j];
        long long ta = pro >> 30;
        ta << 30;
        pro = pro - ta;
      }
  cout << pro;
}
