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 = 1e7;
int p[N + 7], prim[N + 7], cnt;
long long anst, ans, f[N + 7], g[N + 7];
int n;
void init() {
  f[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!prim[i]) {
      p[++cnt] = i;
      prim[i] = i;
      f[i] = i - 1;
      g[i] = 1;
    }
    for (int j = 1; j <= cnt; j++) {
      if (p[j] * i > n) break;
      prim[p[j] * i] = p[j];
      g[p[j]]++;
      if (p[j] == prim[i]) {
        f[p[j] * i] = f[i] * p[j];
        break;
      }
      f[p[j] * i] = f[i] * (p[j] - 1);
    }
  }
  for (int i = 1; i <= n; i++) g[i] += g[i - 1];
}
long long QAQ(int l, int r) {
  if (r < l)
    return 0;
  else
    return g[r] - g[l - 1];
}
int main() {
  scanf("%d", &n);
  init();
  for (int i = 1; i <= n; i++) anst += f[i] - 1;
  ans = 1ll * (n - 1) * (n - 2) / 2 - anst;
  ans += 2 * anst;
  for (int j = 1; j <= cnt && p[j] <= n; j++) {
    long long yy = g[p[j]] - g[p[j] - 1];
    ans +=
        yy * QAQ(max(p[j] + 1, n / p[j] + 1), max(p[j] + 1, (n >> 1) + 1) - 1);
    ans -= 2 * yy * QAQ(max(p[j] + 1, (n >> 1) + 1), n);
  }
  cout << ans << endl;
  return 0;
}