#include <bits/stdc++.h>
const int N = 1.5 * 10000000 + 16, M = 1e5, OO = 0x3f3f3f3f, mod = 1e9 + 7;
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
const double PI = acos(-1.0);
using namespace std;
long long gcd(long long x, long long y) { return (!y) ? x : gcd(y, x % y); }
long long lcm(long long x, long long y) { return ((x / gcd(x, y)) * y); }
void file() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int dp[10000000];
int div2(long long x) {
  set<long long> st;
  if (~dp[x]) return dp[x];
  for (long long i = 1; i * i <= x; i++)
    if (x % i == 0) st.insert(i), st.insert(x / i);
  return dp[x] = st.size();
}
int main() {
  file();
  long long a, b, c;
  cin >> a >> b >> c;
  long long sum = 0;
  memset(dp, -1, sizeof dp);
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++)
      for (int k = 1; k <= c; k++) {
        sum += div2(i * j * k);
        sum %= 1073741824;
      }
  cout << sum << '\n';
}

### Prompt

Your task is to create 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 N = 1.5 * 10000000 + 16, M = 1e5, OO = 0x3f3f3f3f, mod = 1e9 + 7;
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
const double PI = acos(-1.0);
using namespace std;
long long gcd(long long x, long long y) { return (!y) ? x : gcd(y, x % y); }
long long lcm(long long x, long long y) { return ((x / gcd(x, y)) * y); }
void file() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int dp[10000000];
int div2(long long x) {
  set<long long> st;
  if (~dp[x]) return dp[x];
  for (long long i = 1; i * i <= x; i++)
    if (x % i == 0) st.insert(i), st.insert(x / i);
  return dp[x] = st.size();
}
int main() {
  file();
  long long a, b, c;
  cin >> a >> b >> c;
  long long sum = 0;
  memset(dp, -1, sizeof dp);
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++)
      for (int k = 1; k <= c; k++) {
        sum += div2(i * j * k);
        sum %= 1073741824;
      }
  cout << sum << '\n';
}

```