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 maxN = 1e7 + 5;
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
int sieve[maxN];
int phi[maxN];
int numD[maxN];
int ps[maxN];
long long cnt[3];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  iota(phi, phi + maxN, -1);
  iota(sieve, sieve + maxN, 0);
  int n;
  scanf("%d", &n);
  phi[1] = 1;
  sieve[1] = 0;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j * i <= n; j++) {
      if (sieve[i] == i && sieve[j * i] == j * i) sieve[j * i] = i;
      if (j != 1) phi[j * i] -= phi[i];
    }
  }
  numD[1] = 0;
  for (int i = 2; i <= n; i++) {
    int p = i / sieve[i];
    numD[i] = numD[p] + (sieve[p] != sieve[i]);
  }
  long long iso = 1;
  for (int i = n / 2 + 1; i <= n; i++)
    if (sieve[i] == i) iso++;
  cnt[0] = 1ll * (n - iso) * (n - iso - 1) / 2;
  for (int i = 2; i <= n; i++) {
    if (sieve[i] == i && i * 2 > n) continue;
    cnt[1] += i - 1 - phi[i];
  }
  int ncl = 1;
  for (int i = 2; i <= n; i++) {
    if (sieve[i] == i)
      ncl++;
    else
      cnt[2] += phi[i] - ncl + numD[i];
  }
  for (int i = 2; i <= n; i++) {
    if (sieve[i] == i && i * 2 > n) continue;
    ps[sieve[i]]++;
  }
  for (int i = 1; i <= n; i++) ps[i] += ps[i - 1];
  for (int i = 2; i <= n; i++) {
    if (sieve[i] != i || i * 2 > n) continue;
    cnt[2] += ps[n / i];
    for (int j = 1; j * i <= n; j++)
      if (sieve[j * i] <= n / i) cnt[2]--;
  }
  for (int i = 2; i <= n; i++) {
    int p = i / sieve[i];
    if (sieve[p] == p && p != sieve[i] && p != 1) cnt[2]--;
  }
  printf("%lld\n", 3 * cnt[0] - cnt[2] - 2 * cnt[1]);
  return 0;
}