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 = 10000005;
int n, tot, prime[N], phi[N], low[N], s[N];
bool not_prime[N];
void get_prime(int n) {
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!not_prime[i]) prime[++tot] = i, phi[i] = i - 1, low[i] = tot;
    for (int j = 1; j <= tot && i * prime[j] <= n; j++) {
      not_prime[i * prime[j]] = 1;
      low[i * prime[j]] = j;
      if (i % prime[j] == 0) {
        phi[i * prime[j]] = phi[i] * prime[j];
        break;
      }
      phi[i * prime[j]] = phi[i] * (prime[j] - 1);
    }
  }
  for (int i = 2; i <= n; i++) s[low[i]]++;
  while (prime[tot] > n / 2) tot--;
  for (int i = tot - 1; i >= 1; i--) s[i] += s[i + 1];
}
int main() {
  scanf("%d", &n);
  get_prime(n);
  long long ans = 0, sum = 0;
  int now = 0;
  for (int i = 2; i <= n; i++) {
    ans += i - 1 - phi[i];
    if (i <= n / 2 || not_prime[i])
      sum += phi[i] - 1 - now;
    else
      now++;
  }
  ans += sum * 2;
  sum = 0;
  int p = tot, lim = sqrt(n), cnt = 0;
  for (int i = 1; i <= tot; i++) {
    if (prime[i] <= lim) continue;
    while ((long long)prime[p - 1] * prime[i] > (long long)n) p--;
    sum += s[p];
    cnt++;
  }
  ans += sum - (long long)cnt * (cnt + 1) / 2;
  printf("%I64d", ans);
  return 0;
}