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 maxn = 10000010;
int prime[maxn], p[maxn], n, tot, phi[maxn], num[maxn], can;
long long one = 0, two = 0, three = 0, s[maxn];
int main() {
  scanf("%d", &n);
  phi[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!p[i]) {
      p[i] = prime[++tot] = i;
      phi[i] = i - 1;
    }
    for (int j = 1; j <= tot && i * prime[j] <= n; j++) {
      p[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) {
        phi[i * prime[j]] = phi[i] * prime[j];
        break;
      }
      phi[i * prime[j]] = phi[i] * (prime[j] - 1);
    }
  }
  for (int i = 2; i <= n; i++) one += i - 1 - phi[i];
  for (int i = 2; i <= n; i++) num[p[i]]++;
  for (int i = 2; i <= n; i++) s[i] = s[i - 1] + num[i];
  for (int i = 2; i <= n; i++) two += 1ll * num[i] * s[n / i];
  for (int i = 2; i <= n; i++)
    if (1ll * p[i] * p[i] <= n) two--;
  two = two / 2 - one;
  three = 0;
  can = n - 1;
  for (int i = 1; i <= tot; i++)
    if (prime[i] * 2 > n) can--;
  three = 1ll * can * (can - 1) / 2 - two - one;
  printf("%lld", one * 1 + two * 2 + three * 3);
  return 0;
}