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:
#include <bits/stdc++.h> 
//#include <atcoder/all>
#define rep(i, n) for(long long int i = 0; i < n; ++i)
#define _rep(i, m, n) for(long long int i = m; i < n; ++i)
using namespace std;
//using namespace atcoder;
typedef long long ll;
typedef pair<int, int> P;
const ll mod = 1000000007;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
string Yes[2] = {"Yes", "No"};
//typedef modint1000000007 mint;
//typedef modint998244353 mint;
using Graph = vector<vector<int>>;
const int inf = 1e9 + 1;
const ll INF = 1e18 + 1;

/*------------------------------------------------------------------*/

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t; cin >> t;

    while(t--){
        int n; cin >> n;
        vector<pair<int,int>> res;
        if(n > 32){
            for(int i = 32; i < n - 1; i++){
                res.push_back({i + 1, n});
            }
            int m = n;
            while(m > 1){
                res.push_back({n, 32});
                m = (m + 31) / 32;
            }
            n = 32;
        }
        for(int i = 2; i <= n - 2; i++){
            res.push_back({i + 1, n});
        }
        for(int i = 1; i <= 5; i++){
            res.push_back({n, 2});
        }
        cout << res.size() << endl;
        for(auto it : res){
            cout << it.first << " " << it.second << endl;
        }
    }
}