#include <iostream>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cassert>

using namespace std;

struct Card {
  int q, p;
};

typedef vector<Card> VC;
typedef vector<int> VI;

const int MAXT = 20;
const int MAXN = 50000;

int num_cases = MAXT;

//assumes card pictures are from 0 to 2n-1 (n = v.size())
void dump(VC& v) {
  cout << v.size() << endl;
  for (auto c : v)
    cout << c.p+1 << ' ' << c.q+1 << endl;
  --num_cases;
}

void scramble(VC& v) {
  int n = v.size();
  VI c(2*n);

  for (int i = 0; i < 2*n; ++i)
    c[i] = i;

  for (int i = 2*n-1; i > 0; --i)
    swap(c[i], c[rand()%(i+1)]);

  for (int i = 0; i < n; ++i) {
    if (rand()&1) swap(v[i].p, v[i].q);
    v[i].p = c[v[i].p];
    v[i].q = c[v[i].q];
  }
}

//random_yes, generated by assigning the first picture of card i to i
//and the other to a random card, then scrambling
void rand_yes(int n) {
  VC v(n);
  for (int i = 0; i < n; ++i) {
    v[i].p = i;
    v[i].q = rand()%n;
  }

  scramble(v);
  dump(v);
}

//first n-1 cards get random pictures from n-2 pictures, so there is not enough
//last card has two new pictures, so the count is ok
void rand_no(int n) {
  assert(n > 1);
  VC v(n);
  for (int i = 0; i < n-1; ++i) {
    v[i].p = i%(n-2);
    v[i].q = (i+1)%(n-2);
  }
  v[n-1].p = n-1;
  v[n-1].q = n;

  scramble(v);
  dump(v);
}

void full_rand(int n) {
  VC v(n);
  
  //Very interesting threshold. Using 2*n almost always produces
  //"yes" instances. This one seems close to 50/50.
  int mod = 39*n/20;

  for (int i = 0; i < n; ++i) {
    v[i].p = rand()%mod;
    v[i].q = rand()%mod;
  }

  dump(v);
}

int main() {
  cout << num_cases << endl;

  while (num_cases)
    if (num_cases&1) rand_yes(MAXN);
    else rand_no(MAXN);

  return 0;
}