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>
int n, pr[10000010], pcnt, phi[10000010], mp[10000010], ms[10000010];
bool com[10000010];
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    if (!com[i]) pr[pcnt++] = i, phi[i] = i - 1, mp[i] = i;
    for (int j = 0; j < pcnt && 1ll * i * pr[j] <= n; j++) {
      com[i * pr[j]] = 1;
      mp[i * pr[j]] = pr[j];
      if (i % pr[j] == 0) {
        phi[i * pr[j]] = phi[i] * pr[j];
        break;
      }
      phi[i * pr[j]] = phi[i] * (pr[j] - 1);
    }
  }
  int xt = 0;
  for (int i = 2; i <= n; i++)
    if (com[i] || i <= n / 2) xt++, ms[mp[i]]++;
  for (int i = n; i--;) ms[i] += ms[i + 1];
  long long A1 = 0, A2 = 0, A3 = 0;
  for (int i = 2; i <= n; i++) A1 += i - 1 - phi[i];
  int rs = 0;
  for (int i = 2; i <= n; i++)
    if (1ll * i * i > n && i * 2 <= n && !com[i]) {
      A3 += ms[n / i + 1];
      for (int j = i; j <= n; j += i)
        if (mp[j] * 1ll * i > n) A3--;
      rs++;
    }
  A3 -= rs * (rs - 1ll) / 2;
  A2 = xt * (xt - 1ll) / 2 - A1 - A3;
  printf("%lld\n", A1 + A2 * 2 + A3 * 3);
}