Problem: 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]. 

Solution (in CPP):

Solution:
//author: hitch_hiker42;
#include<bits/stdc++.h>
using namespace std;

//solution:
#define int int64_t
#define span(a) begin(a), end(a)

void hike() {
  int n; cin >> n;
  vector<pair<int, int>> a;
  for(int i = 1; i <= n; ++i) {
    if(i == 1 || i == 2 || i == 8 || i == n) continue;
    a.push_back({i, n});
  }
  if(n <= 8) {
    int m = n;
    while(m != 1) {
      a.push_back({n, 2});
      m = (m + 1) / 2;
    }
    cout << a.size() << "\n";
    for(auto& [x, y]: a) cout << x << ' ' << y << "\n";
    return;
  }
  int m = n;
  while(m != 1) {
    a.push_back({n, 8});
    m = (m + 7) / 8;
  }
  for(int i = 1; i <= 3; ++i) a.push_back({8, 2});
  cout << a.size() << "\n";
  for(auto& [x, y]: a) cout << x << ' ' << y << "\n";
}

signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr); cout.tie(nullptr);
  int t = 1; cin >> t;
  while(t--) hike();
  return 0;
} //farewell, until we meet again..