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 + 123;
int m[N], phi[N], cnt[N];
vector<long long> ps;
bool mark[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  long long ans1 = 0, ans2 = 0, ans3 = 0;
  cin >> n;
  for (int i = 2; i <= n / 2; i++)
    if (!mark[i]) {
      cnt[i]++;
      ps.push_back(i);
      for (int j = i + i; j <= n; mark[j] = true, j += i)
        if (!m[j]) {
          m[j] = i;
          cnt[i]++;
        }
    }
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (mark[i]) {
      int d = 1, t = i;
      while (t % m[i] == 0) {
        t /= m[i];
        d *= m[i];
      }
      if (d == i)
        phi[i] = i - i / m[i];
      else
        phi[i] = phi[t] * phi[d];
    } else if (i <= n / 2)
      phi[i] = i - 1;
    if (i <= n / 2 || mark[i]) ans1 += i - 1 - phi[i];
  }
  long long k = 0;
  int j = 0;
  for (int i = ps.size() - 1; i >= 0; i--) {
    while (j < ps.size() && (long long)ps[j] * ps[i] <= n) {
      k += cnt[ps[j]];
      j++;
    }
    long long t = (cnt[ps[i]]) * (k - ((long long)ps[i] * ps[i] <= n ? 1 : 0));
    ans2 += t;
  }
  ans2 /= 2;
  for (int i = 2; i <= n; i++)
    if (mark[i] || i <= n / 2) ans3++;
  ans3 = (ans3 - 1) * (ans3) / 2 - ans2;
  ans2 -= ans1;
  cout << ans1 + ans2 * 2 + ans3 * 3 << endl;
  return 0;
}