#include <bits/stdc++.h>
using namespace std;
int dp[1000005];
int countFactors(int n) {
  int len = (int)(sqrt(n)), l = 0, i;
  for (i = 1; i <= len; ++i) {
    if (n % i == 0) {
      l++;
      if (i != n / i) l++;
    }
  }
  return l;
}
int main() {
  int a, b, c, i, j, k, ans = 0;
  scanf("%d %d %d", &a, &b, &c);
  memset(dp, -1, sizeof(dp));
  for (i = 1; i <= a; ++i) {
    for (j = 1; j <= b; ++j) {
      for (k = 1; k <= c; ++k) {
        if (dp[i * k * j] == -1) dp[i * k * j] = countFactors(i * k * j);
        ans = (ans % 1073741824 + dp[i * k * j] % 1073741824) % 1073741824;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
using namespace std;
int dp[1000005];
int countFactors(int n) {
  int len = (int)(sqrt(n)), l = 0, i;
  for (i = 1; i <= len; ++i) {
    if (n % i == 0) {
      l++;
      if (i != n / i) l++;
    }
  }
  return l;
}
int main() {
  int a, b, c, i, j, k, ans = 0;
  scanf("%d %d %d", &a, &b, &c);
  memset(dp, -1, sizeof(dp));
  for (i = 1; i <= a; ++i) {
    for (j = 1; j <= b; ++j) {
      for (k = 1; k <= c; ++k) {
        if (dp[i * k * j] == -1) dp[i * k * j] = countFactors(i * k * j);
        ans = (ans % 1073741824 + dp[i * k * j] % 1073741824) % 1073741824;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

```