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;
const int maxn = 2000 + 10;
const long long mod = 1073741824;
int a, b, c, n, tot, x, y;
int prime[maxn], miu[maxn], G[maxn][maxn];
bool check[maxn];
int gcd(int a, int b) {
  if (G[a][b]) return G[a][b];
  if (b == 0) return G[a][b] = a;
  return G[a][b] = gcd(b, a % b);
}
void prepare() {
  miu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!check[i]) prime[tot++] = i, miu[i] = -1;
    for (int j = 0; j < tot; j++) {
      if (i * prime[j] > n) break;
      check[i * prime[j]] = true;
      if (i % prime[j] == 0) {
        miu[i * prime[j]] = 0;
        break;
      }
      miu[i * prime[j]] = -miu[i];
    }
  }
}
int main() {
  scanf("%d %d %d", &a, &b, &c);
  n = max(max(a, b), c);
  prepare();
  long long ans = 0;
  for (int d = 1; d <= n; d++)
    if (miu[d] != 0)
      for (int i = 1; i <= a; i++)
        if (gcd(i, d) == 1) {
          x = 0, y = 0;
          for (int j = 1; j <= b / d; j++)
            if (gcd(i, j) == 1) x += b / (j * d);
          for (int k = 1; k <= c / d; k++)
            if (gcd(i, k) == 1) y += c / (k * d);
          ans = (ans + (long long)miu[d] * (long long)x * (long long)y *
                           (long long)(a / i)) %
                mod;
        }
  cout << (ans + mod) % mod << endl;
  return 0;
}