Problem: You are given a positive integer n. Let's build a graph on vertices 1, 2, ..., n in such a way that there is an edge between vertices u and v if and only if <image>. Let d(u, v) be the shortest distance between u and v, or 0 if there is no path between them. Compute the sum of values d(u, v) over all 1 ≤ u < v ≤ n.

The gcd (greatest common divisor) of two positive integers is the maximum positive integer that divides both of the integers.

Input

Single integer n (1 ≤ n ≤ 107).

Output

Print the sum of d(u, v) over all 1 ≤ u < v ≤ n.

Examples

Input

6


Output

8


Input

10


Output

44

Note

All shortest paths in the first example: 

  * <image>
  * <image>
  * <image>
  * <image>
  * <image>
  * <image>



There are no paths between other pairs of vertices.

The total distance is 2 + 1 + 1 + 2 + 1 + 1 = 8.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 10000010;
int n, prime[N], phi[N], s[N], cnt;
bool isnotprime[N];
void get_prime() {
  for (int i = 2; i <= n; i++) {
    if (!isnotprime[i]) prime[++cnt] = i, phi[i] = i - 1, s[i]++;
    for (int j = 1; j <= cnt && i * prime[j] <= n; j++) {
      s[prime[j]]++, isnotprime[i * prime[j]] = true;
      if (i % prime[j])
        phi[i * prime[j]] = phi[i] * (prime[j] - 1);
      else {
        phi[i * prime[j]] = phi[i] * prime[j];
        break;
      }
    }
  }
  for (int i = n; i; i--) s[i] += s[i + 1];
}
int main() {
  scanf("%d", &n);
  get_prime();
  long long sum = n - 1, c1 = 0, c2 = 0, c3 = 0;
  for (int i = 1; i <= cnt; i++)
    if (prime[i] * 2 > n) sum--;
  sum = sum * (sum - 1) / 2;
  for (int i = 2; i <= n; i++) c1 += i - 1 - phi[i];
  c2 = 1ll * (n - 1) * (n - 2) / 2 - c1;
  int tmp = 0;
  for (int i = 1; i <= cnt; i++)
    if (1ll * prime[i] * prime[i] > n) {
      c2 -= s[n / prime[i] + 1], tmp++;
    }
  c2 += tmp, c2 += 1ll * tmp * (tmp - 1) / 2, c3 = sum - c1 - c2;
  printf("%lld\n", c1 + c2 * 2 + c3 * 3);
  return 0;
}