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;
int isNotPrime[1000005];
void sieve() {
  isNotPrime[0] = isNotPrime[1] = 1;
  for (int i = 2; i < 1000005; ++i) {
    if (!isNotPrime[i])
      for (int j = i; j < 1000005; j += i) {
        if (!isNotPrime[j]) {
          isNotPrime[j] = i;
        }
      }
  }
}
int numberOfDivisors[1000005];
void divisors() {
  int e, x, y;
  numberOfDivisors[1] = 1;
  for (int i = 2; i < 1000005; i++) {
    x = i, y = i;
    while (x != 1) {
      e = 0;
      while (x % isNotPrime[y] == 0) {
        e++;
        x /= isNotPrime[y];
      }
      if (numberOfDivisors[i])
        numberOfDivisors[i] *= (e + 1);
      else
        numberOfDivisors[i] = (e + 1);
      y = x;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  sieve();
  divisors();
  long long a, b, c;
  cin >> a >> b >> c;
  long long ans = 0;
  for (int i = 1; i <= a; ++i) {
    for (int j = 1; j <= b; ++j) {
      for (int k = 1; k <= c; ++k) {
        ans = (ans % 1000000007 + numberOfDivisors[i * j * k] % 1000000007) %
              1000000007;
      }
    }
  }
  cout << ans;
  return 0;
}