Problem: Jzzhu has picked n apples from his big apple tree. All the apples are numbered from 1 to n. Now he wants to sell them to an apple store. 

Jzzhu will pack his apples into groups and then sell them. Each group must contain two apples, and the greatest common divisor of numbers of the apples in each group must be greater than 1. Of course, each apple can be part of at most one group.

Jzzhu wonders how to get the maximum possible number of groups. Can you help him?

Input

A single integer n (1 ≤ n ≤ 105), the number of the apples.

Output

The first line must contain a single integer m, representing the maximum number of groups he can get. Each of the next m lines must contain two integers — the numbers of apples in the current group.

If there are several optimal answers you can print any of them.

Examples

Input

6


Output

2
6 3
2 4


Input

9


Output

3
9 3
2 4
6 8


Input

2


Output

0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void print_container(ostream& os, const T& c) {
  const char* _s = " ";
  if (!c.empty()) {
    __typeof__(c.begin()) last = --c.end();
    for (__typeof__((c).begin()) it = (c).begin(); it != (c).end(); ++it) {
      os << *it;
      if (it != last) os << _s;
    }
  }
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& c) {
  print_container(os, c);
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& c) {
  print_container(os, c);
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& c) {
  print_container(os, c);
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const deque<T>& c) {
  print_container(os, c);
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, const map<T, U>& c) {
  print_container(os, c);
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <typename T>
void print(T a, int n, const string& split = " ") {
  for (int i = 0; i < n; i++) {
    cout << a[i];
    if (i + 1 != n) cout << split;
  }
  cout << endl;
}
template <typename T>
void print2d(T a, int w, int h, int width = -1, int br = 0) {
  for (int i = 0; i < h; ++i) {
    for (int j = 0; j < w; ++j) {
      if (width != -1) cout.width(width);
      cout << a[i][j] << ' ';
    }
    cout << endl;
  }
  while (br--) cout << endl;
}
template <typename T>
void input(T& a, int n) {
  for (int i = 0; i < n; ++i) cin >> a[i];
}
template <typename T, typename U>
void upmin(T& a, const U& b) {
  a = min<T>(a, b);
}
template <typename T, typename U>
void upmax(T& a, const U& b) {
  a = max<T>(a, b);
}
template <typename T>
void uniq(T& a) {
  sort(a.begin(), a.end());
  a.erase(unique(a.begin(), a.end()), a.end());
}
template <class T>
string to_s(const T& a) {
  ostringstream os;
  os << a;
  return os.str();
}
template <class T>
T to_T(const string& s) {
  istringstream is(s);
  T res;
  is >> res;
  return res;
}
void fast_io() {
  cin.tie(0);
  ios::sync_with_stdio(false);
}
bool in_rect(int x, int y, int w, int h) {
  return 0 <= x && x < w && 0 <= y && y < h;
}
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
vector<bool> prime_table(int upto) {
  vector<bool> p(upto + 1, true);
  p[0] = p[1] = false;
  for (int i = 2; i * i <= upto; ++i)
    if (p[i])
      for (int j = i + i; j <= upto; j += i) p[j] = false;
  return p;
}
vector<int> primes(int upto, const vector<bool>& p = prime_table(0)) {
  if ((int)p.size() <= upto) return primes(upto, prime_table(upto));
  vector<int> primes;
  for (int i = 0; i <= upto; ++i)
    if (p[i]) primes.push_back(i);
  return primes;
}
int main() {
  int n;
  cin >> n;
  const int M = ((long long)(1e6)) + 810;
  bool used[M] = {};
  vector<int> ps = primes(n);
  reverse((ps).begin(), (ps).end());
  vector<pair<int, int> > res;
  for (int p : ps) {
    static vector<int> a;
    a.clear();
    for (int i = p; i <= n; i += p)
      if (!used[i]) a.push_back(i);
    if (a.size() >= 2 && (a.size() & 1)) a.erase(a.end() - 1 - (a.back() & 1));
    for (int i = 0; i + 1 < (int)a.size(); i += 2) {
      used[a[i]] = used[a[i + 1]] = true;
      res.push_back(pair<int, int>(a[i], a[i + 1]));
    }
  }
  printf("%d\n", (int)res.size());
  for (auto& it : res) printf("%d %d\n", it.first, it.second);
}