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 n, m, tp[10050000], phi[10050000], p[10050000 >> 3], tn, s[10050000];
void sieve() {
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!phi[i]) phi[p[++tn] = tp[i] = i] = i - 1;
    for (int j = 1; j <= tn && i * p[j] <= n; j++) {
      tp[i * p[j]] = p[j];
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      }
      phi[i * p[j]] = phi[i] * (p[j] - 1);
    }
  }
  for (int i = 2; i <= n; i++) s[tp[i]]++;
  for (int i = 1; i <= n; i++) s[i] += s[i - 1];
}
int calc(int l, int r) { return (l >= r) ? 0 : s[r] - s[l]; }
int main() {
  scanf("%d", &n);
  sieve();
  long long perp = 0;
  for (int i = 2; i <= n; i++) perp += phi[i] - 1;
  long long ans1 = 1ll * (n - 1) * (n - 2) / 2 - perp, ans3 = 0, ans4 = 0;
  for (int i = 1; i <= tn; i++) {
    int buf = s[p[i]] - s[p[i] - 1];
    ans3 += 1ll * buf * calc(max(n / p[i], p[i]), n / 2);
    ans4 += 1ll * buf * calc(max(n / 2, p[i]), n);
  }
  printf("%lld", ans1 + (perp - ans3 - ans4) * 2 + ans3 * 3);
  return 0;
}