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;
#define ll long long
#define space " "
#define MOD (long long) 1000000007
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename T> void __f(const char* variable_name, T&& value){cout << variable_name << ": " << value << endl;}
template <typename T1, typename... T2> void __f(const char* names, T1&& arg1, T2&&... args){const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << ": " << arg1 << " |";__f(comma + 1, args...);}

ll n;

void solve(){
  cin >> n;
  vector<pair<ll, ll>> ans;
  while (n > 2) {
    ll c = ceil(sqrt(n));
    for (ll i = c + 1; i < n; i++) ans.push_back({i, i + 1});
    ans.push_back({n, c});
    ans.push_back({n, c});
    n = c;
  }
  cout << ans.size() << endl;
  for (auto i : ans) cout << i.first << ' ' << i.second << endl;
}

int main(){
  ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
  ll t = 1;
  // cout << "\n\nBLAH BLAH BLAH\n\n";
  cin >> t;
  while(t--){
    solve();
  }
  return 0;
}