#include <bits/stdc++.h>
using namespace std;
const int ha = 1 << 30;
int a, b, c, miu[4000000 + 5];
int zs[4000000 / 5], t = 0, D, low[4000000 + 5];
int d[4000000 + 5], to[4000000 + 5], ans = 0;
bool v[4000000 + 5];
inline void init() {
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++) to[i * j]++;
  miu[1] = d[1] = low[1] = 1;
  for (int i = 2; i <= 4000000; i++) {
    if (!v[i]) zs[++t] = i, miu[i] = -1, low[i] = i, d[i] = 2;
    for (int j = 1, u; j <= t && (u = zs[j] * i) <= 4000000; j++) {
      v[u] = 1;
      if (!(i % zs[j])) {
        low[u] = low[i] * zs[j];
        if (low[i] == i)
          d[u] = d[i] + 1;
        else
          d[u] = d[low[u]] * d[i / low[i]];
        break;
      }
      low[u] = zs[j], d[u] = d[i] << 1, miu[u] = -miu[i];
    }
  }
  for (int i = 1; i <= 4000000; i++) d[i] += d[i - 1];
}
inline int add(int x, int y) {
  x += y;
  return x >= ha ? x - ha : x;
}
inline void calc() {
  for (int i = 1, sum; i <= c; i++)
    if (miu[i]) {
      sum = 0;
      for (int j = i; j <= D; j += i)
        sum = add(sum, to[j] * (long long)(d[j / i] - d[j / i - 1]) % ha);
      sum = sum * (long long)d[c / i] % ha;
      if (miu[i] == 1)
        ans = add(ans, sum);
      else
        ans = add(ans, ha - sum);
    }
  printf("%d\n", ans);
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  if (a > b) swap(a, b);
  if (a > c) swap(a, c);
  if (b > c) swap(b, c);
  D = a * b;
  init();
  calc();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int ha = 1 << 30;
int a, b, c, miu[4000000 + 5];
int zs[4000000 / 5], t = 0, D, low[4000000 + 5];
int d[4000000 + 5], to[4000000 + 5], ans = 0;
bool v[4000000 + 5];
inline void init() {
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++) to[i * j]++;
  miu[1] = d[1] = low[1] = 1;
  for (int i = 2; i <= 4000000; i++) {
    if (!v[i]) zs[++t] = i, miu[i] = -1, low[i] = i, d[i] = 2;
    for (int j = 1, u; j <= t && (u = zs[j] * i) <= 4000000; j++) {
      v[u] = 1;
      if (!(i % zs[j])) {
        low[u] = low[i] * zs[j];
        if (low[i] == i)
          d[u] = d[i] + 1;
        else
          d[u] = d[low[u]] * d[i / low[i]];
        break;
      }
      low[u] = zs[j], d[u] = d[i] << 1, miu[u] = -miu[i];
    }
  }
  for (int i = 1; i <= 4000000; i++) d[i] += d[i - 1];
}
inline int add(int x, int y) {
  x += y;
  return x >= ha ? x - ha : x;
}
inline void calc() {
  for (int i = 1, sum; i <= c; i++)
    if (miu[i]) {
      sum = 0;
      for (int j = i; j <= D; j += i)
        sum = add(sum, to[j] * (long long)(d[j / i] - d[j / i - 1]) % ha);
      sum = sum * (long long)d[c / i] % ha;
      if (miu[i] == 1)
        ans = add(ans, sum);
      else
        ans = add(ans, ha - sum);
    }
  printf("%d\n", ans);
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  if (a > b) swap(a, b);
  if (a > c) swap(a, c);
  if (b > c) swap(b, c);
  D = a * b;
  init();
  calc();
  return 0;
}

```