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 INF = 0x3f3f3f3f, maxn = 2005;
int gcd[maxn][maxn], mu[maxn], p[maxn], vis[maxn], res;
void init(int n = 2000) {
  res = 0;
  mu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) p[res++] = i, mu[i] = -1;
    for (int j = 0; j < res && i * p[j] <= n; j++) {
      vis[i * p[j]] = 1;
      if (i % p[j])
        mu[i * p[j]] = -mu[i];
      else {
        mu[i * p[j]] = 0;
        break;
      }
    }
  }
  for (int i = 0; i <= n; i++) gcd[i][0] = gcd[0][i] = i;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++) gcd[j][i] = gcd[i][j] = gcd[j][i % j];
}
int deal(int n, int d, int i) {
  int ans = 0;
  for (int j = 1; j <= n; j++)
    if (gcd[i][j * d] == 1) ans += n / j;
  return ans;
}
int main() {
  init();
  int a, b, c;
  while (~scanf("%d%d%d", &a, &b, &c)) {
    int ans = 0;
    for (int i = 1; i <= a; i++) {
      int res = 0;
      for (int d = 1; d <= min(b, c); d++) {
        res += ((long long)mu[d] * deal(b / d, d, i) * deal(c / d, d, i) %
                    1073741824 +
                1073741824) %
               1073741824;
        if (res >= 1073741824) res -= 1073741824;
      }
      ans += (long long)(a / i) * res % 1073741824;
      if (ans >= 1073741824) ans -= 1073741824;
    }
    printf("%d\n", ans);
  }
  return 0;
}