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;
template <class T>
bool checkMin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool checkMax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
const int INF = 1e9;
const int MN = 101;
const int MOD = 1073741824;
int d[MN * MN * MN];
int a, b, c;
int N = MN * MN * MN;
int main() {
  memset(d, 0, sizeof(d));
  scanf("%d %d %d", &a, &b, &c);
  int n = a * b * c, m = sqrt(N);
  for (int i = (1); i < (N); i++) {
    for (int j = i; j < N; j += i) d[j]++;
  }
  int sum = 0;
  for (int i = (1); i <= (a); i++)
    for (int j = (1); j <= (b); j++)
      for (int k = (1); k <= (c); k++) sum = (sum + d[i * j * k]) % MOD;
  printf("%d\n", sum);
}
