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 read() {
  char c = getchar();
  long long f = 1, x = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  return f * x;
}
long long n, phi[10000005], p[10000005], prime[1000005], cnt, ans1, ans2, ans3,
    c[10000005];
int main() {
  n = read();
  phi[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!p[i]) {
      p[i] = i;
      phi[i] = i - 1;
      prime[++cnt] = i;
    }
    for (int j = 1; j <= cnt; ++j) {
      if (prime[j] * i > n) break;
      p[i * prime[j]] = prime[j];
      if (!(i % prime[j])) {
        phi[i * prime[j]] = phi[i] * prime[j];
        break;
      } else
        phi[i * prime[j]] = phi[i] * (prime[j] - 1);
    }
  }
  for (int i = 2; i <= n; ++i) ans1 += i - 1 - phi[i], ++c[p[i]];
  for (int i = 2; i <= n; ++i) c[i] = c[i - 1] + c[i];
  for (int i = 2; i <= n; ++i) ans2 += (c[i] - c[i - 1]) * c[n / i];
  for (int i = 2; i <= n; ++i)
    if (p[i] * p[i] <= n) --ans2;
  ans2 = ans2 / 2 - ans1;
  long long cnt_not_prime = n - 1;
  for (int i = 1; i <= cnt; ++i)
    if (prime[i] * 2 > n) --cnt_not_prime;
  ans3 = cnt_not_prime * (cnt_not_prime - 1) / 2 - ans1 - ans2;
  printf("%lld", ans1 + ans2 * 2 + ans3 * 3);
  return 0;
}