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 = 1e7 + 5;
struct bits {
  int limits;
  int c[maxn];
  int lowbit(int x) { return x & -x; }
  bits(int _l = 0) : limits(_l) {}
  void init(int a) { limits = a; }
  void add(int x, int d) {
    for (int i = x; i < limits; i += lowbit(i)) {
      c[i] += d;
    }
  }
  int sum(int x) {
    int ret = 0;
    for (int i = x; i > 0; i -= lowbit(i)) {
      ret += c[i];
    }
    return ret;
  }
} sol;
int n;
int lp[maxn];
long long cnt[maxn];
void solve() {
  cin >> n;
  sol.init(n);
  for (int i = 2; i <= n; i++) {
    if (lp[i] == 0) {
      for (int j = i; j <= n; j += i) {
        if (lp[j] == 0) lp[j] = i;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    cnt[i] = n / i;
    cnt[i] = cnt[i] * (cnt[i] - 1) / 2;
  }
  for (int i = n; i >= 1; i--) {
    for (int j = i + i; j <= n; j += i) {
      cnt[i] -= cnt[j];
    }
  }
  long long ans = 0;
  for (int i = 2; i <= n; i++) {
    ans -= cnt[i];
  }
  for (int i = 2; i <= n; i++) {
    sol.add(lp[i], 1);
  }
  for (int i = 2; i <= n; i++) {
    long long cnt2, cnt3;
    sol.add(lp[i], -1);
    cnt2 = sol.sum(n / lp[i]);
    cnt3 = sol.sum(n / 2);
    cnt3 -= cnt2;
    ans += 2 * cnt2;
    if (2 * lp[i] <= n) ans += 3 * cnt3;
  }
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}