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 + 10;
int phi[N], fen[N], f[N];
bool pr[N];
void upd(int p, int vl) {
  for (p++; p < N; p += p & -p) fen[p] += vl;
  return;
}
int get(int p) {
  int res = 0;
  for (; p > 0; p -= p & -p) res += fen[p];
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  long long ans = 0, sum = 0;
  for (int i = 1; i <= n; i++) {
    phi[i] += i;
    if (!pr[i]) f[i] = i;
    for (int j = 2 * i; j <= n; j += i) {
      phi[j] -= phi[i];
      if (i - 1) pr[j] = true;
      if (f[j] == 0 && i > 1) f[j] = i;
    }
    if (i - 1) sum += phi[i];
  }
  int r = 1;
  for (int i = 2; i <= n; i++) {
    if (!pr[i] && 2 * i > n) r++;
  }
  sum -= 1LL * r * (r - 1) / 2 + 1LL * r * (n - r);
  sum = 1LL * (n - r) * (n - r - 1) / 2 - sum;
  ans += 1LL * (n - r) * (n - r - 1) / 2;
  long long k = 0;
  for (int i = 2; i <= n; i++) {
    int p = f[i];
    k += get(n / p + 1);
    upd(p, 1);
  }
  ans += k - sum;
  ans += 2LL * (1LL * (n - r) * (n - r - 1) / 2 - k);
  cout << ans << endl;
}