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;
const int MAXP = 4e3 + 5;
int n, tot;
int pri[MAXN], phi[MAXN], sum[MAXN];
int f[MAXN], pc[MAXN], cnt[MAXN];
bool bo[MAXN];
void init() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    if (!bo[i]) {
      pri[tot++] = i;
      phi[i] = i - 1;
      f[i] = i;
      pc[i] = 1;
    }
    cnt[i] = cnt[i - 1] + (1 ^ bo[i]);
    sum[f[i]]++;
    for (int j = 0; j < tot && i * pri[j] <= n; ++j) {
      bo[i * pri[j]] = 1;
      f[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) {
        phi[i * pri[j]] = phi[i] * pri[j];
        pc[i * pri[j]] = pc[i];
        break;
      }
      phi[i * pri[j]] = phi[i] * (pri[j] - 1);
      pc[i * pri[j]] = pc[i] + 1;
    }
  }
  for (int i = 2; i <= n; ++i) sum[i] += sum[i - 1];
}
void solve() {
  long long now = 1ll * sum[n / 2] * (sum[n / 2] - 1) / 2;
  long long ans = 0;
  for (int i = 2; i <= n; ++i) {
    if (bo[i]) {
      ans += i - phi[i] - 1;
      now -= i - phi[i] - 1;
      ans += 2ll * (phi[i] - cnt[i] + pc[i] - 1);
      now -= phi[i] - cnt[i] + pc[i] - 1;
    }
  }
  for (int i = 2; i <= n; ++i) {
    if (!bo[i]) {
      ans += 2ll * (sum[n / i] - n / i);
      now -= (sum[n / i] - n / i);
      ans -= 2ll * cnt[min(i - 1, n / i)];
      now += cnt[min(i - 1, n / i)];
      if (i > n / i) {
        ans += 2ll * (sum[i] - sum[i - 1]);
        now -= sum[i] - sum[i - 1];
      }
    }
  }
  ans += 3ll * now;
  printf("%I64d\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}