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 lpf[10001000], mob[10001000], lgt[10001000];
int prm[10001000], pcnt;
int main() {
  int n;
  long long ans = 0;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int tmp = n / i;
    if (!lpf[i]) {
      prm[pcnt++] = i;
      lpf[i] = i;
      mob[i] = 1;
      tmp--;
    } else
      lgt[lpf[i]]++;
    ans -= 1LL * mob[i] * tmp * (tmp - 1) / 2;
    for (int j = 0; j < pcnt && prm[j] <= n / i; j++) {
      lpf[i * prm[j]] = prm[j];
      if (i % prm[j] == 0) break;
      mob[i * prm[j]] = -mob[i];
    }
  }
  ans += 1LL * (n - 1 - pcnt) * (n - 2 - pcnt);
  for (int i = n - 1; i >= 1; i--) {
    lgt[i] += lgt[i + 1];
  }
  for (int i = 0; i < pcnt && prm[i] * 2 <= n; i++) {
    ans += (n - 1 - pcnt) * 2;
    ans -= (n / prm[i] - 1);
    ans += lgt[n / prm[i] + 1];
  }
  while (pcnt > 0 && prm[pcnt - 1] * 2 > n) pcnt--;
  ans += 1LL * pcnt * (pcnt - 1) / 2 * 3;
  int j = 0;
  for (int i = pcnt - 1; i >= 0; i--) {
    while (j < pcnt && 1LL * prm[i] * prm[j] <= n) j++;
    ans -= min(j, i);
  }
  printf("%lld\n", ans);
  return 0;
}