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;
int n, bo[10000005], minp[10000005], m, pri[10000005];
long long ans, phi[10000005], sum[10000005];
long long find(int l, int r) {
  if (l > r) return 0;
  return sum[r] - sum[l - 1];
}
int main() {
  scanf("%d", &n);
  bo[1] = phi[1] = 1;
  for (long long i = 2; i <= n; i++) {
    if (!bo[i]) {
      pri[++m] = i;
      minp[i] = i;
      phi[i] = i - 1;
    }
    for (long long j = 1; j <= m && pri[j] * i <= n; j++) {
      bo[i * pri[j]] = 1;
      minp[i * pri[j]] = 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);
    }
  }
  for (int i = 2; i <= n; i++) ++sum[minp[i]];
  for (int i = 2; i <= n; i++) sum[i] += sum[i - 1];
  long long t = 0;
  for (int i = 2; i <= n; i++) t += phi[i] - 1;
  ans = t + (((long long)(n - 1) * (n - 2)) >> 1);
  for (int i = 1; i <= m; i++) {
    long long t = sum[pri[i]] - sum[pri[i] - 1];
    ans += t * (find(max(pri[i], n / pri[i]) + 1, max(pri[i], n >> 1)) -
                (find(max(pri[i], n >> 1) + 1, n) << 1));
  }
  printf("%lld", ans);
}