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 N = 10000007;
int p[N], mi[N], c[N], v[N], tot, n;
long long phi[N], ans;
int rd() {
  char c = getchar();
  int s = 0, f = 1;
  while (c > '9' || c < '0') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    s = (s << 1) + (s << 3) + (c ^ 48), c = getchar();
  return s * f;
}
void prepare() {
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!v[i]) {
      p[++tot] = i, mi[i] = i, phi[i] = i - 1;
    }
    for (int j = 1; j <= tot; j++) {
      if (p[j] * i > n) break;
      v[i * p[j]] = 1;
      mi[i * p[j]] = p[j];
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      } else
        phi[i * p[j]] = phi[i] * (p[j] - 1);
    }
    phi[i] += phi[i - 1];
  }
  for (int i = 1; i <= n; i++) c[mi[i]]++;
  for (int i = 1; i <= n; i++) c[i] += c[i - 1];
}
int get(int l, int r) {
  if (r < l) return 0;
  return c[r] - c[l - 1];
}
int main() {
  n = rd();
  prepare();
  ans = 1ll * (n - 1) * (n - 2) / 2 + phi[n] - n;
  for (int i = 1; i <= tot; i++) {
    int cnt = c[p[i]] - c[p[i] - 1];
    ans += 1ll *
           get(max(p[i] + 1, n / p[i] + 1), max(p[i] + 1, n / 2 + 1) - 1) * cnt;
    ans -= 2ll * get(max(p[i] + 1, (n >> 1) + 1), n) * cnt;
  }
  printf("%lld\n", ans);
  return 0;
}