#include <bits/stdc++.h>
using namespace std;
long long arr[1000005] = {0}, arr2[1000005] = {0};
int main() {
  int a, b, c, ch = 0;
  long long co = 0, p = 0, x = 0;
  scanf("%d%d%d", &a, &b, &c);
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        arr[co] = i * j * k;
        co++;
      }
    }
  }
  co = 0;
  while (arr[co]) {
    if (arr2[arr[co]]) {
      x += arr2[arr[co]];
      co++;
      continue;
    }
    for (int i = 1; i * i <= arr[co]; i++) {
      if (arr[co] % i == 0) p++;
      if (i * i == arr[co]) {
        ch = 1;
      }
    }
    p = p * 2;
    if (ch) {
      p--;
      ch = 0;
    }
    arr2[arr[co]] = p;
    x += p;
    p = 0;
    co++;
  }
  printf("%llu", x);
  return 0;
}

### Prompt

Create a solution in CPP for 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>
using namespace std;
long long arr[1000005] = {0}, arr2[1000005] = {0};
int main() {
  int a, b, c, ch = 0;
  long long co = 0, p = 0, x = 0;
  scanf("%d%d%d", &a, &b, &c);
  for (int i = 1; i <= a; i++) {
    for (int j = 1; j <= b; j++) {
      for (int k = 1; k <= c; k++) {
        arr[co] = i * j * k;
        co++;
      }
    }
  }
  co = 0;
  while (arr[co]) {
    if (arr2[arr[co]]) {
      x += arr2[arr[co]];
      co++;
      continue;
    }
    for (int i = 1; i * i <= arr[co]; i++) {
      if (arr[co] % i == 0) p++;
      if (i * i == arr[co]) {
        ch = 1;
      }
    }
    p = p * 2;
    if (ch) {
      p--;
      ch = 0;
    }
    arr2[arr[co]] = p;
    x += p;
    p = 0;
    co++;
  }
  printf("%llu", x);
  return 0;
}

```