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 = 10000005;
using ll = long long;
int n, cnt[N], pr[N], ptr, d[N];
ll phi[N];
bool npr[N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!npr[i]) pr[++ptr] = i, phi[i] = i - 1, d[i] = i;
    for (int j = 1; j <= ptr && i * pr[j] <= n; j++) {
      npr[i * pr[j]] = 1;
      d[i * pr[j]] = pr[j];
      if (i % pr[j])
        phi[i * pr[j]] = phi[i] * (pr[j] - 1);
      else {
        phi[i * pr[j]] = phi[i] * pr[j];
        break;
      }
    }
    ++cnt[d[i]];
  }
  for (int i = 2; i <= n; i++) cnt[i] += cnt[i - 1];
  ll psum = accumulate(phi + 1, phi + n + 1, 0ll) - n;
  ll ans = ((n - 1ll) * (n - 2) / 2 - psum) + psum * 2;
  for (int i = 1; i <= ptr; i++) {
    ll cc = cnt[pr[i]] - cnt[pr[i] - 1];
    ans +=
        cc * (cnt[max(pr[i], n / 2)] - cnt[max(pr[i] + 1, n / pr[i] + 1) - 1]);
    ans -= 2 * cc * (cnt[n] - cnt[max(pr[i] + 1, n / 2 + 1) - 1]);
  }
  cout << ans << endl;
  return 0;
}