#include <bits/stdc++.h>
const int MN = 2e3 + 10;
int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
int A, B, C, mu[MN], l[MN], p[MN], P;
unsigned int f;
bool c[MN], ok[MN][MN];
unsigned int solve(int x, int m) {
  unsigned int r = 0;
  for (int i = 1; i <= m; ++i)
    if (ok[i][x]) r += m / i;
  return r;
}
int main() {
  mu[1] = 1;
  for (int i = 2; i < MN; ++i) {
    if (!c[i]) l[i] = P, p[P++] = i, mu[i] = -1;
    for (int j = 0, k; (k = i * p[j]) < MN; ++j) {
      c[k] = 1;
      l[k] = j;
      if (l[i] == j) break;
      mu[k] = -mu[i];
    }
  }
  for (int i = 1; i < MN; ++i)
    for (int j = 1; j < MN; ++j) ok[i][j] = gcd(i, j) == 1;
  scanf("%d%d%d", &A, &B, &C);
  for (int g = 1; g < MN; ++g)
    if (mu[g])
      for (int i = 1; i <= A; ++i)
        if (ok[i][g]) f += mu[g] * (A / i) * solve(i, B / g) * solve(i, C / g);
  printf("%d\n", f & 0x3fffffff);
  return 0;
}

### Prompt

Please formulate 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 MN = 2e3 + 10;
int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
int A, B, C, mu[MN], l[MN], p[MN], P;
unsigned int f;
bool c[MN], ok[MN][MN];
unsigned int solve(int x, int m) {
  unsigned int r = 0;
  for (int i = 1; i <= m; ++i)
    if (ok[i][x]) r += m / i;
  return r;
}
int main() {
  mu[1] = 1;
  for (int i = 2; i < MN; ++i) {
    if (!c[i]) l[i] = P, p[P++] = i, mu[i] = -1;
    for (int j = 0, k; (k = i * p[j]) < MN; ++j) {
      c[k] = 1;
      l[k] = j;
      if (l[i] == j) break;
      mu[k] = -mu[i];
    }
  }
  for (int i = 1; i < MN; ++i)
    for (int j = 1; j < MN; ++j) ok[i][j] = gcd(i, j) == 1;
  scanf("%d%d%d", &A, &B, &C);
  for (int g = 1; g < MN; ++g)
    if (mu[g])
      for (int i = 1; i <= A; ++i)
        if (ok[i][g]) f += mu[g] * (A / i) * solve(i, B / g) * solve(i, C / g);
  printf("%d\n", f & 0x3fffffff);
  return 0;
}

```