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 LL = long long;
const int N = 1e7 + 2;
int phi[N], sm[N], n, tot, pri[N / 12], pz[N];
LL t0 = 1, t1, t2, t3;
bool vis[N];
int main() {
  scanf("%d", &n);
  phi[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!vis[i]) pri[++tot] = i, phi[i] = i - 1, sm[i] = i, t0 += i * 2 > n;
    for (int j = 1; j <= tot && i * pri[j] <= n; ++j) {
      vis[i * pri[j]] = 1;
      sm[i * pri[j]] = std::min(sm[i], pri[j]);
      if (i % pri[j] == 0) {
        phi[i * pri[j]] = phi[i] * pri[j];
        break;
      } else
        phi[i * pri[j]] = phi[i] * (pri[j] - 1);
    }
    ++pz[sm[i]];
  }
  for (int i = 2; i <= n; ++i) t1 += i - 1 - phi[i], pz[i] += pz[i - 1];
  for (int i = 2; i <= n; ++i) t2 += pz[n / sm[i]];
  t1 *= 2;
  t2 -= t1;
  for (int i = 2; i <= n; ++i)
    if ((LL)sm[i] * sm[i] <= n) --t2;
  t3 = (LL)n * (n - 1) / 2 - t0 * (n - t0) - t0 * (t0 - 1LL) / 2 - t1 / 2 -
       t2 / 2;
  t3 *= 2;
  printf("%lld\n", t1 + t2 * 2 + t3 * 3 >> 1);
  return 0;
}