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 = (int)1e7 + 10;
int cnt[N];
int phi[N];
int low[N];
int cur[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  phi[1] = 0;
  for (int i = 1; i <= n; i++) {
    low[i] = i;
    cnt[i] = 1;
  }
  for (int i = 2; i <= n; i++) {
    if (low[i] == i) {
      phi[i] = i - 1;
    } else {
      phi[i] = phi[i / low[i]];
      if ((i / low[i]) % low[i] == 0)
        phi[i] *= low[i];
      else
        phi[i] *= (low[i] - 1);
    }
    for (int j = i; j <= n; j += i) {
      if (low[j] == j) low[j] = i;
      cnt[j]++;
    }
  }
  long long one = 0;
  for (int i = 1; i <= n; i++) {
    one += (i - 1 - phi[i]);
  }
  long long two = 0;
  for (int i = 2; i <= n; i++) {
    cur[low[i]]++;
  }
  for (int i = 2; i <= n; i++) cur[i] += cur[i - 1];
  int lim;
  for (int i = 2; i <= n; i++) {
    lim = n / low[i];
    two += cur[lim];
    if (low[i] * 1ll * low[i] <= n) two--;
  }
  two -= 2 * one;
  two /= 2;
  long long three = n * 1ll * (n - 1) / 2ll;
  three -= one;
  three -= two;
  long long cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (i == 1 || (low[i] == i && i * 2 > n)) cnt++;
  }
  three -= cnt * 1ll * (n - cnt);
  three -= cnt * 1ll * (cnt - 1) / 2ll;
  cout << one + two * 2ll + 3ll * three << "\n";
  return 0;
}