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;
long long f[2005][2005], g[2005][2005];
const int p = 1 << 30;
int a, b, c;
int gcd(int x, int y) {
  if (!y) return x;
  return gcd(y, x % y);
}
long long F(int x, int y) {
  int n = b / y;
  long long ans = 0;
  for (int j = 1; j <= n; ++j)
    if (gcd(x, j) == 1) ans = (ans + n / j) % p;
  return ans;
}
long long G(int x, int y) {
  int n = c / y;
  long long ans = 0;
  for (int j = 1; j <= n; ++j)
    if (gcd(x, j) == 1) ans = (ans + n / j) % p;
  return ans;
}
int miu[2005], v[2005], cnt, pp[2005];
void init() {
  miu[1] = 1;
  for (int i = 2; i <= 2000; ++i) {
    if (!v[i]) {
      pp[++cnt] = i;
      miu[i] = -1;
    }
    for (int j = 1; j <= cnt && i * pp[j] <= 2000; ++j) {
      v[i * pp[j]] = 1;
      if (i % pp[j] == 0) break;
      miu[i * pp[j]] = -miu[i];
    }
  }
}
int main() {
  init();
  scanf("%d%d%d", &a, &b, &c);
  if (a > b) swap(a, b);
  if (a > c) swap(a, c);
  if (b > c) swap(b, c);
  for (int i = 1; i <= a; ++i)
    for (int j = 1; j <= b; ++j) f[i][j] = F(i, j);
  for (int i = 1; i <= a; ++i)
    for (int j = 1; j <= c; ++j) g[i][j] = G(i, j);
  long long ans = 0;
  for (int i = 1; i <= a; i++) {
    for (int q = 1; q <= b; ++q)
      if (gcd(i, q) == 1)
        ans =
            (ans + 1ll * (a / i) % p * miu[q] % p * f[i][q] % p * g[i][q] % p) %
            p;
  }
  printf("%I64d\n", (ans + p) % p);
  return 0;
}