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 N = 1e7 + 5, M = 4005;
int n, tot, p[N], phi[N], sum[N], f[N], pc[N], cnt[N];
long long ans, now;
bool bz[N];
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    if (!bz[i]) p[tot++] = i, phi[i] = i - 1, f[i] = i, pc[i] = 1;
    cnt[i] = cnt[i - 1] + (1 ^ bz[i]);
    sum[f[i]]++;
    for (int j = 0; j < tot && i * p[j] <= n; j++) {
      bz[i * p[j]] = 1;
      f[i * p[j]] = p[j];
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        pc[i * p[j]] = pc[i];
        break;
      }
      phi[i * p[j]] = phi[i] * (p[j] - 1);
      pc[i * p[j]] = pc[i] + 1;
    }
  }
  for (int i = 2; i <= n; i++) sum[i] += sum[i - 1];
  now = 1ll * sum[n / 2] * (sum[n / 2] - 1) / 2;
  for (int i = 2; i <= n; i++)
    if (bz[i]) {
      ans += i - phi[i] - 1;
      now -= i - phi[i] - 1;
      ans += 2 * (phi[i] - cnt[i] + pc[i] - 1);
      now -= phi[i] - cnt[i] + pc[i] - 1;
    }
  for (int i = 2; i <= n / 2; i++)
    if (!bz[i]) {
      ans += 2 * (sum[n / i] - n / i);
      now -= sum[n / i] - n / i;
      ans -= 2 * cnt[min(i - 1, n / i)];
      now += cnt[min(i - 1, n / i)];
      if (i > n / i)
        ans += 2 * (sum[i] - sum[i - 1]), now -= sum[i] - sum[i - 1];
    }
  ans += 3 * now;
  printf("%I64d\n", ans);
  return 0;
}