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:

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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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void solve();
vector<long long> seive(1001, 1);
vector<long long> prime;
int32_t main() {
  long long t = 1;
  for (long long i = 2; i < 1001; i++) {
    if (seive[i]) {
      prime.push_back(i);
      for (long long p = i * i; p < 1001; p += i) {
        seive[p] = 0;
      }
    }
  }
  while (t--) {
    solve();
  }
}
long long d(long long n) {
  long long total = 1;
  for (long long x : prime) {
    if (x > n) {
      break;
    }
    long long cnt = 0;
    while (n % x == 0) {
      n /= x;
      cnt++;
    }
    total *= (cnt + 1);
  }
  return total;
}
void solve() {
  long long a, b, c;
  cin >> a >> b >> c;
  long long arr[a * b * c + 1];
  arr[1] = 1;
  arr[0] = 0;
  for (long long num = 2; num <= a * b * c; num++) {
    arr[num] = d(num);
  }
  long long ans = 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++) {
        ans += arr[i * j * k];
      }
    }
  }
  cout << ans << "\n";
}