You have an array a_1, a_2, ..., a_n where a_i = i.

In one step, you can choose two indices x and y (x ≠ y) and set a_x = \left⌈ (a_x)/(a_y) \right⌉ (ceiling function).

Your goal is to make array a consist of n - 1 ones and 1 two in no more than n + 5 steps. Note that you don't have to minimize the number of steps.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first and only line of each test case contains the single integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the length of array a.

It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print the sequence of operations that will make a as n - 1 ones and 1 two in the following format: firstly, print one integer m (m ≤ n + 5) — the number of operations; next print m pairs of integers x and y (1 ≤ x, y ≤ n; x ≠ y) (x may be greater or less than y) — the indices of the corresponding operation.

It can be proven that for the given constraints it's always possible to find a correct sequence of operations.

Example

Input


2
3
4


Output


2
3 2
3 2
3
3 4
4 2
4 2

Note

In the first test case, you have array a = [1, 2, 3]. For example, you can do the following: 

  1. choose 3, 2: a_3 = \left⌈ (a_3)/(a_2) \right⌉ = 2 and array a = [1, 2, 2]; 
  2. choose 3, 2: a_3 = \left⌈ 2/2 \right⌉ = 1 and array a = [1, 2, 1]. 

You've got array with 2 ones and 1 two in 2 steps.

In the second test case, a = [1, 2, 3, 4]. For example, you can do the following: 

  1. choose 3, 4: a_3 = \left⌈ 3/4 \right⌉ = 1 and array a = [1, 2, 1, 4]; 
  2. choose 4, 2: a_4 = \left⌈ 4/2 \right⌉ = 2 and array a = [1, 2, 1, 2]; 
  3. choose 4, 2: a_4 = \left⌈ 2/2 \right⌉ = 1 and array a = [1, 2, 1, 1]. 
#include <bits/stdc++.h>

using namespace std;

#ifdef BOI
#include "debug.h"
#else
#define debug(args...)
#endif

int main() {
      ios_base :: sync_with_stdio(0);
      cin.tie(0);

      int t; cin >> t;
      while (t--) {
            int n; cin >> n;

            vector <pair <int, int>> ans;

            // if (n == 3) {
            //       cout << "2\n3 2\n3 2\n";
            //       continue;
            // }

            if (n <= (1LL << 8)) {
                  for (int i = 3; i <= n - 1; ++i) {
                        ans.push_back({i, i + 1});
                  }
                  int dd = n;
                  while (dd != 1) {
                        ans.push_back({n, 2});
                        dd = (dd + 1) / 2;
                  }

                  cout << (int) ans.size() << '\n'; 
                  for (auto x : ans) cout << x.first << ' ' << x.second << '\n';
            } else {
                  int dd = 64;

                  for (int i = 3; i < 64; ++i) {
                        ans.push_back({i, i + 1});
                  }

                  for (int i = dd + 1; i <= n - 1; ++i) {
                        ans.push_back({i, i + 1});
                  }

                  int ff = n;
                  while (ff != 1) {
                        ans.push_back({n, 64});
                        ff = (ff + 63) / 64;
                  }
                  // am I fucking retarded wow
                  // I legit thought of this but didnt fucking do it

                  // n - 4 + log128(n) + log2(128) -> I guess it's still a bit more than n + 5
                  // let's try 64

                  while (dd != 1) {
                        ans.push_back({64, 2});
                        dd /= 2;
                  }

                  cout << (int) ans.size() << '\n';
                  for (auto& x : ans) cout << x.first << ' ' << x.second << '\n';
            }

            // fucking retard

      }      
}
