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 div(int n);
using namespace std;
map<int, int> di;
int main(int argc, char *argv[]) {
  int m = 1073741824;
  int a, b, c, i, j, k;
  long long int cnt = 0;
  scanf("%d", &a), scanf("%d", &b), scanf("%d", &c);
  for (i = 1; i <= a; i++) {
    for (j = 1; j <= b; j++) {
      for (k = 1; k <= c; k++) {
        cnt = (cnt + div(i * j * k)) % m;
      }
    }
  }
  printf("%I64d\n", cnt);
  return 0;
}
int div(int n) {
  if (di[n] > 0) {
    return di[n];
  }
  int i = 1, cnt = 0;
  int limit = (int)sqrt(n);
  for (i = 1; i <= limit; i++) {
    if (n % i == 0) {
      cnt = cnt + 2;
    }
  }
  if ((limit * limit) == n) cnt--;
  di[n] = cnt;
  return cnt;
}
