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 + 10;
int n, cnt, phi[N], mu[N], p[N / 10], vis[N], T[N], pre[N];
long long sum1, sum2, sum3, ans, all;
void sieve(int n) {
  phi[1] = mu[1] = 1;
  for (int i = (2); i <= (n); i++) {
    if (!vis[i]) p[++cnt] = vis[i] = i, phi[i] = i - 1, mu[i] = -1;
    for (int j = 1; j <= cnt && i * p[j] <= n; j++) {
      vis[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);
      mu[i * p[j]] = -mu[i];
    }
  }
}
int main() {
  scanf("%d", &n);
  sieve(n);
  for (int i = (2); i <= (n); i++)
    if (vis[i] != i || i <= n / 2) all++;
  all = all * (all - 1) / 2;
  for (int i = (2); i <= (n); i++) sum1 += i - 1 - phi[i];
  ans += sum1;
  for (int i = (2); i <= (n); i++)
    if (vis[i] != i || i <= n / 2) T[vis[i]]++;
  for (int i = (2); i <= (n); i++) pre[i] = pre[i - 1] + T[i];
  for (int i = (2); i <= (n); i++)
    if (vis[i] != i || i <= n / 2) sum2 += pre[n / vis[i]];
  int m = sqrt(n);
  for (int i = (2); i <= (n); i++) {
    long long tmp = 0;
    tmp += (long long)(n / i) * (n / i);
    if (i > m && vis[i] == i) tmp--;
    sum2 += mu[i] * tmp;
  }
  sum2 /= 2;
  ans += sum2 * 2;
  ans += (all - sum1 - sum2) * 3;
  printf("%lld\n", ans);
  return 0;
}