#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long mod_exp(long long x, long long y, long long mm) {
  if (y == 0)
    return (1);
  else if (y % 2 == 0)
    return (mod_exp((x * x) % mm, y / 2, mm));
  else
    return ((x * mod_exp((x * x) % mm, (y - 1) / 2, mm)) % mm);
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
long long divisors(long long n) {
  if (n == 1) return 1;
  long long cnt = 0;
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else {
        cnt += 2;
      }
    }
  }
  return cnt + 2;
}
long long solve() {
  long long a, b, c;
  cin >> a >> b >> c;
  vector<long long> v(a * b * c + 1);
  for (long long i = 1; i <= a * b * c; i++) {
    for (long long j = i; j <= a * b * c; j += i) v[j]++;
  }
  long long res = 0;
  for (long long i = 1; i <= a; i++) {
    for (long long j = 1; j <= b; j++) {
      for (long long k = 1; k <= c; k++) {
        long long x = i * j * k;
        res = (res % 1073741824 + v[x] % 1073741824) % 1073741824;
      }
    }
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << solve();
  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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long mod_exp(long long x, long long y, long long mm) {
  if (y == 0)
    return (1);
  else if (y % 2 == 0)
    return (mod_exp((x * x) % mm, y / 2, mm));
  else
    return ((x * mod_exp((x * x) % mm, (y - 1) / 2, mm)) % mm);
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
long long divisors(long long n) {
  if (n == 1) return 1;
  long long cnt = 0;
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else {
        cnt += 2;
      }
    }
  }
  return cnt + 2;
}
long long solve() {
  long long a, b, c;
  cin >> a >> b >> c;
  vector<long long> v(a * b * c + 1);
  for (long long i = 1; i <= a * b * c; i++) {
    for (long long j = i; j <= a * b * c; j += i) v[j]++;
  }
  long long res = 0;
  for (long long i = 1; i <= a; i++) {
    for (long long j = 1; j <= b; j++) {
      for (long long k = 1; k <= c; k++) {
        long long x = i * j * k;
        res = (res % 1073741824 + v[x] % 1073741824) % 1073741824;
      }
    }
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << solve();
  return 0;
}

```