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 = 10000000 + 77;
int n, t = 1;
int F[N], Phi[N], Mnp[N];
long long A[4];
int main() {
  scanf("%d", &n);
  Phi[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (Mnp[i] > 0) continue;
    for (int j = i; j <= n; j += i)
      if (Mnp[j] == 0) Mnp[j] = i;
  }
  A[0] = n - 1;
  for (int i = 2; i <= n; ++i) {
    ++F[Mnp[i]];
    if (Mnp[i / Mnp[i]] == Mnp[i])
      Phi[i] = Phi[i / Mnp[i]] * Mnp[i];
    else
      Phi[i] = Phi[i / Mnp[i]] * (Mnp[i] - 1);
    A[1] += i - Phi[i] - 1;
    if (Mnp[i] == i && i > (n >> 1)) A[0] += n - ++t;
  }
  for (int i = 2; i <= n; ++i) F[i] += F[i - 1];
  for (int i = 2; i <= n; ++i) {
    if (Mnp[i] != i || i * 1ll * i <= n || i > (n >> 1)) continue;
    int le = n / i;
    A[3] += F[i] - F[le] - 1;
  }
  A[2] = n * 1ll * (n - 1) / 2 - A[0] - A[1] - A[3];
  printf("%lld\n", A[1] + 2 * A[2] + 3 * A[3]);
  return 0;
}