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, Phi[10000005], Prim[10000005], tot, num[10000005];
int p1[10000005], s1[10000005];
long long an1, an2, an3, t1;
inline void Init() {
  Phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!p1[i]) p1[i] = Prim[++tot] = i, Phi[i] = i - 1;
    for (int j = 1; i * Prim[j] <= n && j <= tot; j++) {
      p1[i * Prim[j]] = Prim[j];
      if (i % Prim[j] == 0) {
        Phi[i * Prim[j]] = Phi[i] * Prim[j];
        break;
      }
      Phi[i * Prim[j]] = Phi[i] * (Prim[j] - 1);
    }
  }
}
int main() {
  scanf("%d", &n);
  Init();
  for (int i = 1; i <= tot; i++)
    if (Prim[i] * 2 > n) t1++;
  for (int i = 2; i <= n; i++) {
    an1 += i - 1 - Phi[i];
    num[p1[i]]++;
  }
  for (int i = 2; i <= n; i++) s1[i] = s1[i - 1] + num[i];
  for (int i = 2; i <= n; i++) {
    an2 += 1ll * num[i] * s1[n / i];
    if (1ll * p1[i] * p1[i] <= n) an2--;
  }
  an2 = an2 / 2 - an1;
  an3 = 1ll * (n - 1 - t1) * (n - 2 - t1) / 2 - an1 - an2;
  printf("%lld", an1 + an2 * 2 + an3 * 3);
}