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;
int n, prime[10000005], p[10000005], phi[10000005], c[10000005], pt;
int main() {
  scanf("%d", &n);
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!p[i]) prime[++pt] = i, phi[i] = i - 1, c[pt] = 1;
    for (int j = 1; j <= pt && i * prime[j] <= n; j++) {
      int u = i * prime[j];
      p[i * prime[j]] = 1;
      c[j]++;
      if (i % prime[j] == 0) {
        phi[u] = phi[i] * prime[j];
        break;
      }
      phi[u] = phi[i] * (prime[j] - 1);
    }
  }
  long long ans0, ans1, ans2, ans3;
  int ct = 1, ct2;
  for (int i = 1; i <= pt; i++)
    if (prime[i] * 2 > n) ct++;
  ans0 = 2ll * ct * (n - 1) - 1ll * (ct - 1) * ct;
  ans1 = 0;
  for (int i = 1; i <= n; i++) ans1 += phi[i];
  ans1 = (1ll * n * (n - 1) / 2 + 1 - ans1) * 2;
  ans3 = 0;
  int nw = pt - ct + 1;
  ct2 = 0;
  for (int i = 1; i <= pt - ct + 1; i++) {
    while (nw > i && 1ll * prime[i] * prime[nw] > n) {
      ct2 += c[nw];
      nw--;
    }
    while (nw < i) {
      nw++;
      ct2 -= c[nw];
    }
    ans3 += 1ll * ct2 * c[i];
  }
  ans3 *= 2;
  ans2 = 1ll * n * (n - 1) - ans0 - ans1 - ans3;
  long long ans = ans1 + 2 * ans2 + 3 * ans3;
  printf("%lld\n", ans / 2);
  return 0;
}