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 = 1e7;
int n, NBb;
int EmnP[N + 3], Ephi[N + 3], EdivP[N + 3], ECmnP[N + 3];
bool Bb[N + 3];
void Process() {
  for (int i = 1; i <= n; i++) Ephi[i] = i;
  for (int i = 2; i <= n; i++) {
    if (EmnP[i]) continue;
    for (int j = i; j <= n; j += i) {
      if (!EmnP[j]) EmnP[j] = i;
      Ephi[j] /= i;
      Ephi[j] *= i - 1;
    }
  }
  for (int i = 1; i <= n; i++)
    if (i == 1 || (EmnP[i] == i && i * 2 > n)) Bb[i] = true, NBb++;
  for (int i = 2; i <= n; i++)
    if (EmnP[i] == i)
      EdivP[i] = 1;
    else if (EmnP[i / EmnP[i]] == EmnP[i])
      EdivP[i] = EdivP[i / EmnP[i]];
    else
      EdivP[i] = EdivP[i / EmnP[i]] + 1;
}
long long GetDis1() {
  long long Rr = 0;
  for (int i = 2; i <= n; i++)
    if (!Bb[i]) Rr += i - 1 - Ephi[i];
  return Rr;
}
long long GetDis2() {
  long long Rr = 0;
  int Nprm = 1;
  for (int i = 2; i <= n; i++)
    if (EmnP[i] == i)
      Nprm++;
    else
      Rr = Rr + Ephi[i] - Nprm + EdivP[i];
  for (int i = 1; i <= n; i++)
    if (!Bb[i]) ECmnP[EmnP[i]]++;
  for (int i = 1; i <= n; i++) ECmnP[i] += ECmnP[i - 1];
  for (int i = 2; i <= n; i++) {
    if (EmnP[i] != i || Bb[i]) continue;
    Rr += ECmnP[n / i];
    for (int j = i; j <= n; j += i)
      if (EmnP[j] <= n / i) Rr--;
  }
  for (int i = 2; i <= n; i++) {
    int Va = EmnP[i];
    int Vb = i / Va;
    if (Vb != 1 && EmnP[Vb] == Vb && Va != Vb) Rr--;
  }
  return Rr;
}
int main() {
  scanf("%d", &n);
  Process();
  long long Cc1 = GetDis1(), Cc2 = GetDis2(),
            Cc3 = (long long)(n - NBb) * (n - NBb - 1) / 2 - Cc1 - Cc2;
  printf("%lld\n", Cc1 + Cc2 * 2 + Cc3 * 3);
  return 0;
}