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, p[1000000], a[10000005], ph[10000005];
long long ans, tmp;
bool used[10000005];
int main() {
  scanf("%d", &n), ph[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!used[i]) p[++m] = i, ph[i] = i - 1, a[i]++;
    tmp += ph[i] - 1;
    for (int j = 1, x = i << 1; j <= m && x <= n; j++, x = i * p[j]) {
      used[x] = 1, a[p[j]]++;
      if (i % p[j])
        ph[x] = ph[i] * (p[j] - 1);
      else {
        ph[x] = ph[i] * p[j];
        break;
      }
    }
  }
  ans = ((long long)(n - 1) * (n - 2) >> 1) + tmp;
  for (int i = 2; i <= n; i++) a[i] += a[i - 1];
  for (int i = 1; i <= m; i++)
    ans += (long long)(a[p[i]] - a[p[i - 1]]) *
           (a[max(p[i] + 1, (n >> 1) + 1) - 1] -
            a[max(p[i] + 1, n / p[i] + 1) - 1] -
            (a[n] - a[max(p[i] + 1, (n >> 1) + 1) - 1] << 1));
  printf("%lld", ans);
  return 0;
}