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 prime[(10000005)], phi[(10000005)], pr[(10000005)], ps[(10000005)],
    sum[(10000005)], n, cnt;
long long tot, cnt0, cnt1, cnt2;
int main() {
  cin >> n, tot = 1LL * (n - 1) * (n - 2) >> 1;
  for (register int i = 2; i <= n; ++i) {
    if (!ps[i]) prime[++cnt] = pr[i] = i, ps[i] = 1, phi[i] = i - 1;
    sum[i] = cnt;
    for (register int j = 1, k; j <= cnt; ++j) {
      k = i * prime[j];
      if (k > n) break;
      pr[k] = prime[j], ps[k] = ps[i];
      if (i % prime[j])
        ++ps[k], phi[k] = phi[i] * (prime[j] - 1);
      else {
        phi[k] = phi[i] * prime[j];
        break;
      }
    }
  }
  for (register int i = 1; i <= cnt; ++i)
    if (prime[i] > n >> 1) cnt0 += n - cnt + i - 2;
  for (register int i = 2; i * i <= n && i <= n >> 1; ++i)
    if (pr[i] == i) cnt2 += sum[min(n >> 1, n / i)] - sum[i];
  for (register int i = 2; i <= n; ++i) {
    cnt1 += i - 1 - phi[i];
    if (pr[i] != i) cnt2 += sum[min(n >> 1, n / pr[i])] - sum[i] + phi[i] - 1;
  }
  cout << cnt1 + 2 * cnt2 + 3 * (tot - cnt0 - cnt1 - cnt2);
  return 0;
}