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 MAX_N = 1e7 + 5;
long long pairc(long long n) { return (n * (n - 1)) / 2; }
int factor[MAX_N];
int sieve[MAX_N];
int cspf[MAX_N];
int ccspf[MAX_N];
int mobius[MAX_N];
int main() {
  long long n;
  cin >> n;
  vector<long long> primes;
  for (long long i = 2; i <= n; i++) {
    if (!sieve[i]) {
      primes.push_back(i);
      for (long long j = i; j <= n; j += i) {
        factor[j] = i;
        if (!sieve[j]) {
          sieve[j] = 1;
          cspf[i]++;
        }
      }
    }
  }
  for (long long i = 2; i <= n; i++) {
    if (factor[i] == i) {
      mobius[i] = 1;
    } else if (factor[i / factor[i]] == factor[i]) {
      mobius[i] = 0;
    } else {
      mobius[i] = -mobius[i / factor[i]];
    }
  }
  long long isol = 0;
  for (long long i = 2; i <= n; i++) {
    if (factor[i] != i || 2 * i <= n) {
      isol++;
    } else {
      cspf[i]--;
    }
  }
  ccspf[n] = cspf[n];
  for (int i = n - 1; i >= 0; i--) {
    ccspf[i] = cspf[i] + ccspf[i + 1];
  }
  long long ans = isol * (isol - 1);
  long long bigp = 0;
  for (int p : primes) {
    if (2 * p <= n) {
      int s = n / p + 1;
      if (s <= p) s = p + 1;
      bigp += (long long)cspf[p] * (long long)ccspf[s];
    }
  }
  ans += bigp;
  long long smallp = 0;
  for (long long i = 2; i <= n; i++) {
    smallp += (long long)mobius[i] * pairc(n / i);
  }
  ans -= smallp;
  cout << ans << endl;
}