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;
int phi[10000001], leastPrime[10000001], d[10000001];
long long s[10000001];
int main() {
  int n;
  long long ans1, ans2, ans;
  cin >> n;
  for (int i = 1; i <= n; ++i) phi[i] = i;
  ans1 = 0;
  for (int i = 2; i <= n; ++i) {
    if (leastPrime[i] == 0) {
      int k = 0;
      for (int j = i; j <= n; j += i) {
        if (leastPrime[j] == 0) {
          ++k;
          leastPrime[j] = i;
        }
        phi[j] /= i;
        phi[j] *= (i - 1);
      }
      d[i] = k;
    }
    s[i] = s[i - 1] + d[i];
    ans1 += (i - 1 - phi[i]);
  }
  ans2 = 0;
  for (int i = 2; i <= n; ++i) {
    if (d[i]) {
      ans2 += d[i] * s[n / i];
      if (1ll * i * i <= n) ans2 -= d[i];
    }
  }
  ans2 /= 2;
  long long ans3 = s[n / 2] * (s[n / 2] - 1) / 2;
  ans3 -= ans2;
  ans2 -= ans1;
  ans = ans1 + 2 * ans2 + 3 * ans3;
  cout << ans << '\n';
  return 0;
}