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>
//memset(,0, sizeof );
using namespace std;
#define fastIO ios::sync_with_stdio(0), ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
typedef long long ll;
typedef long int li;
typedef long double ld;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<vi> vvi;
typedef vector<vd> vvd;
typedef vector<string> vs;
typedef pair<int, int> pi;
const int N = 1e9;
const int mod = 1e9 + 7;
string S = " ", E = "\n";

int main() {
    fastIO

    int t;
    cin >> t;
    while (t--) {
        vector<pair<int ,int>>p;
        int n,steps=0,k=0;
        cin>>n;
        int n2=n;
        while(n2>2){
            n2 = ceil(sqrt(n2));
            for(int i=n2+1;i<n;i++){
                p.push_back(make_pair(i,n));
                steps++;
            }
            p.push_back(make_pair(n,n2));
            p.push_back(make_pair(n,n2));
            n = n2 ;
            steps+=2;
        }
        cout<<steps<<E;
        
        for(auto it:p)cout<<it.first<<S<<it.second<<E;

    }

    return 0;
}