#include <bits/stdc++.h>
using namespace std;
const int N = 10000000;
int mu[N + 10], prime[N], f[N + 10], g[N + 10];
bool vis[N + 10];
int n, tot;
long long ans, sum;
int main() {
  scanf("%d", &n);
  mu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) prime[++tot] = i, f[i] = i, mu[i] = -1;
    for (int j = 1; j <= tot && i * prime[j] <= n; j++) {
      vis[i * prime[j]] = 1;
      f[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
    g[f[i]]++;
  }
  while (tot && prime[tot] > n / 2) tot--;
  int sum1 = 0, sum2 = 0, now = tot;
  for (int i = 1; i <= tot; i++) sum1 += g[prime[i]];
  for (int i = 1; i <= tot; i++) {
    while (now > i && prime[i] * prime[now] > n)
      sum2 += g[prime[now]], sum1 -= g[prime[now]], now--;
    if (now >= i) sum1 -= g[prime[i]];
    if (now < i) sum2 -= g[prime[i]];
    ans += 1ll * g[prime[i]] * (g[prime[i]] - 1) + 2ll * g[prime[i]] * sum1 +
           3ll * g[prime[i]] * sum2;
  }
  sum = 1ll * n * (n - 1);
  for (int i = 1; i <= n; i++) sum -= 1ll * mu[i] * (n / i) * (n / i);
  sum++;
  ans -= sum / 2;
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 10000000;
int mu[N + 10], prime[N], f[N + 10], g[N + 10];
bool vis[N + 10];
int n, tot;
long long ans, sum;
int main() {
  scanf("%d", &n);
  mu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) prime[++tot] = i, f[i] = i, mu[i] = -1;
    for (int j = 1; j <= tot && i * prime[j] <= n; j++) {
      vis[i * prime[j]] = 1;
      f[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
    g[f[i]]++;
  }
  while (tot && prime[tot] > n / 2) tot--;
  int sum1 = 0, sum2 = 0, now = tot;
  for (int i = 1; i <= tot; i++) sum1 += g[prime[i]];
  for (int i = 1; i <= tot; i++) {
    while (now > i && prime[i] * prime[now] > n)
      sum2 += g[prime[now]], sum1 -= g[prime[now]], now--;
    if (now >= i) sum1 -= g[prime[i]];
    if (now < i) sum2 -= g[prime[i]];
    ans += 1ll * g[prime[i]] * (g[prime[i]] - 1) + 2ll * g[prime[i]] * sum1 +
           3ll * g[prime[i]] * sum2;
  }
  sum = 1ll * n * (n - 1);
  for (int i = 1; i <= n; i++) sum -= 1ll * mu[i] * (n / i) * (n / i);
  sum++;
  ans -= sum / 2;
  printf("%I64d\n", ans);
  return 0;
}

```