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, pri[2000100], phi[10010000], tax[10010000], sum[10010000], tot = 0,
                                                                  low[10010000];
bool nopri[10010000];
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while ((ch < '0') && (ch > '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ((ch >= '0') && (ch <= '9')) {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * f;
}
void sieve(int N) {
  phi[1] = 1;
  for (register int i = 2; i <= N; i++) {
    if (!nopri[i]) {
      pri[++tot] = i;
      phi[i] = i - 1;
      low[i] = i;
    }
    int j;
    for (j = 1; j <= tot && pri[j] * i <= N; j++) {
      nopri[i * pri[j]] = 1;
      low[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 (register int i = 2; i <= N; i++) tax[low[i]]++;
  for (register int i = 1; i <= N; i++) sum[i] = sum[i - 1] + tax[i];
}
int main() {
  n = read();
  sieve(n);
  long long ans = 0, ans1 = 0, ans2 = 0, ans3 = 0;
  for (register int i = 2; i <= n; i++) ans1 += (i - phi[i] - 1);
  for (register int i = 2; i <= n; i++) ans2 += 1ll * tax[i] * sum[n / i];
  for (register int i = 2; i <= n; i++)
    if (1ll * low[i] * low[i] <= n) ans2--;
  ans2 = ans2 / 2 - ans1;
  int m = n - 1;
  for (register int i = tot; i >= 1; i--)
    if (pri[i] * 2 > n)
      m--;
    else
      break;
  ans3 = 1ll * m * (m - 1) / 2 - ans1 - ans2;
  ans = ans1 + ans2 * 2 + ans3 * 3;
  printf("%lld\n", ans);
  return 0;
}