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 = 4000010, maxm = 2010;
long long s[maxn], f[maxn], P[maxm], Q[maxm];
int a, b, c, p[maxm], vis[maxm], mu[maxm];
void inc(long long &a, long long b) {
  a += b;
  if (a > (1ll << 30)) a -= (1ll << 30);
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  mu[1] = 1;
  for (int i = 2; i <= c; i++) {
    if (!vis[i]) p[++p[0]] = i, mu[i] = -1;
    for (int j = 1; j <= p[0] && i * p[j] <= c; j++) {
      vis[i * p[j]] = 1;
      if (i % p[j] == 0) {
        mu[i * p[j]] = 0;
        break;
      } else
        mu[i * p[j]] = -mu[i];
    }
  }
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++) f[i * j]++;
  for (int i = 1; i <= a * b; i++)
    for (int j = 1; j <= a * b / i; j++) inc(s[i], f[i * j]);
  for (int i = 1; i <= c; i++)
    for (int j = 1; j <= a * b / i; j++) inc(P[i], s[i * j]);
  for (int i = 1; i <= c; i++)
    for (int j = 1; j <= i; j++) inc(Q[i], i / j);
  long long ans = 0;
  for (int i = 1; i <= c; i++)
    ans = (ans + mu[i] * P[i] * Q[c / i] % (1ll << 30)) % (1ll << 30);
  printf("%I64d", (ans + (1ll << 30)) % (1ll << 30));
  return 0;
}