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>
const int N = 10000005;
int n, md[N], pr[N / 10], pt, phi[N];
void sieve(int n) {
  phi[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!md[i]) pr[++pt] = i, md[i] = i, phi[i] = i - 1;
    for (int j = 1, t; j <= pt && pr[j] <= md[i]; ++j) {
      if ((t = i * pr[j]) > n) break;
      md[t] = pr[j];
      if (pr[j] == md[i]) {
        phi[t] = phi[i] * pr[j];
        break;
      }
      phi[t] = phi[i] * (pr[j] - 1);
    }
  }
}
long long c(long long n) { return n * (n - 1) / 2; }
int cnt[N];
long long pre[N];
int main() {
  std::ios::sync_with_stdio(0), std::cin.tie(0);
  std::cin >> n, sieve(n);
  long long c0 = 1, c1 = c(n) + 1, c2 = 0, c3 = 0;
  for (int i = 1; i <= pt; ++i) c0 += 2 * pr[i] > n;
  c0 = c(c0) + c0 * (n - c0);
  for (int i = 1; i <= n; ++i) c1 -= phi[i];
  for (int i = 2; i <= n; ++i) ++cnt[md[i]];
  for (int i = 2; i <= n; ++i) pre[i] = pre[i - 1] + cnt[i];
  for (int i = 1; i <= pt && pr[i] * pr[i] <= n; ++i)
    c2 += c(cnt[pr[i]]) + cnt[pr[i]] * (pre[n / pr[i]] - pre[pr[i]]);
  c2 -= c1, c3 = c(n) - c0 - c1 - c2;
  std::cout << c1 + c2 * 2 + c3 * 3 << '\n';
  return 0;
}