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 long long Mod = 1ll << 30;
int a, b, c, pr[2010UL], g[2010UL][2010UL], u[2010UL];
bool vis[2010UL];
int Gcd(int x, int y) {
  if (g[x][y]) return g[x][y];
  if (y == 0) return g[x][y] = x;
  return g[x][y] = Gcd(y, x % y);
}
void Pre(int r) {
  u[1] = 1;
  for (int i = 2; i <= r; ++i) {
    if (!vis[i]) {
      pr[++pr[0]] = i;
      u[i] = -1;
    }
    for (int j = 1; j <= pr[0] && 1ll * i * pr[j] <= r; ++j) {
      int v = i * pr[j];
      vis[v] = true;
      if (!(i % pr[j])) {
        u[v] = 0;
        break;
      } else
        u[v] = -u[i];
    }
  }
  return;
}
long long Cal(int x, int r) {
  long long ret = 0;
  for (int i = 1; i <= r; ++i)
    if (Gcd(x, i) == 1) ret += r / i;
  return ret;
}
int main() {
  Pre(2000);
  scanf("%d%d%d", &a, &b, &c);
  long long ans = 0;
  for (int i = 1; i <= a; ++i) {
    for (int j = 1, r = min(b, c); j <= r; ++j) {
      if (Gcd(i, j) == 1) {
        ans += 1ll * (a / i) * u[j] * Cal(i, b / j) * Cal(i, c / j);
        while (ans > Mod) ans -= Mod;
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}