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;
long long n, ans;
int isp[10000000 + 5], p[10000000 + 5], phi[10000000 + 5], cnt,
    lp[10000000 + 5];
long long c[10000000 + 5];
void Prime() {
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!isp[i]) {
      p[++cnt] = i;
      phi[i] = i - 1;
      lp[i] = i;
    }
    for (int j = 1; j <= cnt; j++) {
      if (i * p[j] > n) break;
      isp[i * p[j]] = 1;
      lp[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++) c[lp[i]]++;
  for (int i = 1; i <= n; i++) c[i] += c[i - 1];
}
long long sum(int l, int r) {
  if (r < l) return 0;
  return c[r] - c[l - 1];
}
int main() {
  scanf("%lld", &n);
  Prime();
  for (int i = 2; i <= n; i++) ans += (phi[i] - 1);
  ans += ((n - 1) * (n - 2)) / 2LL;
  for (int i = 1; i <= cnt; i++) {
    long long tp = c[p[i]] - c[p[i] - 1];
    ans += tp * sum(max(p[i] + 1, (int)n / p[i] + 1), max(p[i], (int)n / 2));
    ans -= 2 * tp * sum(max(p[i] + 1, (int)n / 2 + 1), (int)n);
  }
  printf("%lld", ans);
}