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 N = 2005;
int mu[N], prime[N], rec[N][N];
bool pd[N];
inline void make_mu() {
  int sz = 0;
  mu[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!pd[i]) {
      mu[i] = -1;
      prime[++sz] = i;
    }
    for (int j = 1; j <= sz; j++) {
      if (i * prime[j] >= N) break;
      pd[i * prime[j]] = 1;
      if (i % prime[j] == 0)
        mu[i * prime[j]] = 0;
      else
        mu[i * prime[j]] = -mu[i];
    }
  }
}
inline int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
inline int getgcd(int x, int y) {
  if (rec[x][y]) return rec[x][y];
  rec[x][y] = gcd(x, y);
  return rec[x][y];
}
inline int cal(int ub, int base, int k) {
  int ret = 0;
  for (int i = base; i <= ub; i += base)
    if (getgcd(i, k) == 1) ret += (ub / i);
  return ret;
}
int main() {
  make_mu();
  int a, b, c;
  long long ans = 0;
  scanf("%d%d%d", &a, &b, &c);
  for (int i = 1; i <= c; i++)
    for (int d = 1; d <= a; d++)
      if (mu[d] != 0)
        ans = (ans + (long long)(c / i) * mu[d] * cal(a, d, i) * cal(b, d, i)) %
              1073741824ll;
  ans = (ans + 1073741824ll) % 1073741824ll;
  printf("%I64d\n", ans);
}