#include <bits/stdc++.h>
const int INF = 0x3f3f3f3f;
using namespace std;
const int N = 1e6;
long long sum[N];
int i, j, k;
int a, b, c;
long long ans;
void Find() {
  for (int i = 1; i <= N; i++)
    for (int j = i; j <= N; j += i) sum[j]++;
}
int main() {
  long long int n;
  Find();
  while (~scanf("%d %d %d", &a, &b, &c)) {
    ans = 0;
    for (i = 1; i <= a; i++) {
      for (j = 1; j <= b; j++) {
        for (k = 1; k <= c; k++) {
          ans += sum[i * j * k];
        }
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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:

<image>

Find the sum modulo 1073741824 (230).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 2000).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Examples

Input

2 2 2


Output

20


Input

4 4 4


Output

328


Input

10 10 10


Output

11536

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.

### Response

```cpp
#include <bits/stdc++.h>
const int INF = 0x3f3f3f3f;
using namespace std;
const int N = 1e6;
long long sum[N];
int i, j, k;
int a, b, c;
long long ans;
void Find() {
  for (int i = 1; i <= N; i++)
    for (int j = i; j <= N; j += i) sum[j]++;
}
int main() {
  long long int n;
  Find();
  while (~scanf("%d %d %d", &a, &b, &c)) {
    ans = 0;
    for (i = 1; i <= a; i++) {
      for (j = 1; j <= b; j++) {
        for (k = 1; k <= c; k++) {
          ans += sum[i * j * k];
        }
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```