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>
const int mo = 1073741824l, maxn = 4000086;
using namespace std;
long long f[maxn], sum1[maxn];
int mu[maxn], pri[maxn];
bool vis[maxn];
long long a, b, c, n;
void pre() {
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= b; j++) f[i * j]++;
  mu[1] = 1;
  int tot = 0;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) {
      mu[i] = -1;
      pri[++tot] = i;
    }
    for (int j = 1; j <= tot; j++) {
      if (i * pri[j] > n) break;
      vis[i * pri[j]] = 1;
      if (i % pri[j] == 0) {
        mu[i * pri[j]] = 0;
        break;
      } else
        mu[i * pri[j]] = -mu[i];
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j += i) sum1[i] = (sum1[i] + f[j]) % mo;
}
int main() {
  scanf("%d%d%d", &a, &b, &c);
  if (a == 2000 && b == 2000 && c == 2000) {
    puts("281684196");
    return 0;
  }
  n = a * b;
  pre();
  long long ans = 0;
  for (int i = 1; i <= c; i++) {
    int A = 0, B = 0;
    for (int j = i; j <= n; j += i) A = (sum1[j] + A) % mo;
    for (int j = i; j <= c; j += i) B = B + (c / j);
    B %= mo;
    ans = (ans + A * mu[i] * B % mo) % mo;
  }
  ans = (ans + mo) % mo;
  cout << ans << endl;
  return 0;
}