#include <bits/stdc++.h>
using namespace std;
inline void Rd(int &res) {
  static char c;
  res = 0;
  while ((c = getchar()) < 48)
    ;
  do res = (res << 1) + (res << 3) + (c ^ 48);
  while ((c = getchar()) > 47);
}
const int maxn = 2e3 + 100, P = 1 << 30;
bool cur1;
int mo[maxn];
int prime[maxn], tot;
bool mark[maxn];
void init() {
  mo[1] = 1;
  for (int i = 2; i < maxn; i++) {
    if (!mark[i]) {
      prime[++tot] = i;
      mo[i] = -1;
    }
    for (int j = 1; j <= tot; j++) {
      int t = prime[j] * i;
      if (t >= maxn) break;
      mo[t] = -mo[i];
      mark[t] = 1;
      if (i % prime[j] == 0) {
        mo[t] = 0;
        break;
      }
    }
  }
  for (int i = 0; i < maxn; i++)
    if (mo[i] < 0) mo[i] = P - 1;
}
int g[maxn][maxn];
int gcd(int a, int b) {
  return g[a][b] != -1 ? g[a][b] : g[a][b] = (!b ? a : gcd(b, a % b));
}
int solve(int a, int b, int c) {
  int ans = 0;
  for (int x = 1; x <= a; x++) {
    int res = 0;
    for (int d = 1, ed = min(b, c); d <= ed; d++) {
      int tmp1 = 0;
      for (int y = d; y <= b; y += d) {
        if (gcd(x, y) == 1) tmp1 = (tmp1 + b / y) % P;
      }
      int tmp2 = 0;
      for (int z = d; z <= c; z += d) {
        if (gcd(x, z) == 1) tmp2 = (tmp2 + c / z) % P;
      }
      res = (res + 1LL * mo[d] * tmp1 % P * tmp2) % P;
    }
    ans = (ans + 1LL * res * (a / x)) % P;
  }
  return ans;
}
bool cur2;
int main() {
  memset(g, -1, sizeof(g));
  init();
  int a, b, c;
  cin >> a >> b >> c;
  cout << solve(a, b, c) << endl;
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
inline void Rd(int &res) {
  static char c;
  res = 0;
  while ((c = getchar()) < 48)
    ;
  do res = (res << 1) + (res << 3) + (c ^ 48);
  while ((c = getchar()) > 47);
}
const int maxn = 2e3 + 100, P = 1 << 30;
bool cur1;
int mo[maxn];
int prime[maxn], tot;
bool mark[maxn];
void init() {
  mo[1] = 1;
  for (int i = 2; i < maxn; i++) {
    if (!mark[i]) {
      prime[++tot] = i;
      mo[i] = -1;
    }
    for (int j = 1; j <= tot; j++) {
      int t = prime[j] * i;
      if (t >= maxn) break;
      mo[t] = -mo[i];
      mark[t] = 1;
      if (i % prime[j] == 0) {
        mo[t] = 0;
        break;
      }
    }
  }
  for (int i = 0; i < maxn; i++)
    if (mo[i] < 0) mo[i] = P - 1;
}
int g[maxn][maxn];
int gcd(int a, int b) {
  return g[a][b] != -1 ? g[a][b] : g[a][b] = (!b ? a : gcd(b, a % b));
}
int solve(int a, int b, int c) {
  int ans = 0;
  for (int x = 1; x <= a; x++) {
    int res = 0;
    for (int d = 1, ed = min(b, c); d <= ed; d++) {
      int tmp1 = 0;
      for (int y = d; y <= b; y += d) {
        if (gcd(x, y) == 1) tmp1 = (tmp1 + b / y) % P;
      }
      int tmp2 = 0;
      for (int z = d; z <= c; z += d) {
        if (gcd(x, z) == 1) tmp2 = (tmp2 + c / z) % P;
      }
      res = (res + 1LL * mo[d] * tmp1 % P * tmp2) % P;
    }
    ans = (ans + 1LL * res * (a / x)) % P;
  }
  return ans;
}
bool cur2;
int main() {
  memset(g, -1, sizeof(g));
  init();
  int a, b, c;
  cin >> a >> b >> c;
  cout << solve(a, b, c) << endl;
  return 0;
}

```