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;
namespace FGF {
int n;
const int N = 1e7 + 5;
long long ans;
int phi[N], p[N], cnt, mi[N], sum[N], c[N];
void init() {
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!phi[i]) phi[i] = i - 1, p[++cnt] = i, mi[i] = i;
    for (int j = 1; j <= cnt && p[j] * i <= n; j++) {
      mi[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 = 1; i <= n; i++) c[mi[i]]++;
  for (int i = 1; i <= n; i++) sum[i] = c[i] + sum[i - 1];
}
void work() {
  scanf("%d", &n);
  init();
  for (int i = 2; i <= n; i++) ans += i - 1 - phi[i], ans += 2 * (phi[i] - 1);
  for (int i = 1; i <= cnt; i++)
    ans += 1ll * c[p[i]] *
           (max(sum[n / 2] - sum[max(p[i], n / p[i])], 0) -
            max(sum[n] - sum[max(p[i], n / 2)], 0) * 2ll);
  printf("%lld", ans);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}