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;
const long long mx = 1e6 + 3;
#pragma GCC optimize("O3")
vector<long long> divs(mx, 0);
void make() {
  for (long long n = 1; n < mx; n++) {
    for (long long k = n; k < mx; k += n) divs[k]++;
  }
}
int main() {
  make();
  long long x, y, z;
  scanf("%lld", &x);
  scanf("%lld", &y), scanf("%lld", &z);
  long long res = 0;
  long long mod = (1 << 30);
  for (long long n = 1; n <= x; n++) {
    for (long long k = 1; k <= y; k++) {
      for (long long p = 1; p <= z; p++) {
        long long cur = n * k * p;
        res += divs[cur];
        res %= mod;
      }
    }
  }
  printf("%lld", res);
}
