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 dp[1000009] = {0};
const int MOD = 1073741824;
int diva(int x) {
  if (dp[x] != -1) return dp[x];
  if (x <= 2) return x;
  int ret = 0;
  int sqroot = (int)sqrt(x) + 1;
  for (int i = 1; i < sqroot; i++) {
    if (x % i == 0 && i * i == x) {
      ret++;
    } else if (x % i == 0 && i * i != x) {
      ret += 2;
    }
  }
  dp[x] = ret % MOD;
  return ret % MOD;
}
int main() {
  int a, b, c, ans = 0;
  cin >> a >> b >> c;
  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++) {
        ans += diva(i * j * k);
      }
    }
  }
  cout << ans << endl;
  return 0;
}