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>
#define ll long long
#define pii pair<int,int>
using namespace std;
const int mxn=1e5+5;
const int inf=1e9;
vector<pii> res;
int N;
void add(int i,int j) {
	if(res.size() == N + 5) {
		exit(0);
	}
	res.push_back({i,j});
}
void pr() {
	cout<<res.size()<<endl;
	for(pii x : res) cout<<x.first<<" "<<x.second<<endl;
}
void solve() {
    int n;
    cin>>n;
    N = n;
	res.clear();


	if(n < 8) {
		for(int i = 1 ; i < n; i++) {
			if(i != 2) add(i,n);
		}

		while(n != 1) {
			n = (n + 1) / 2;
			res.push_back({N,2});
		}

		
		pr();
		return;
	}

	for(int i = 1; i < n; i++) {
		if(i == 1 || i == 2 || i == 4 || i == 8) continue;
		add(i,n);
	}


	if(n != 8) {
	    int N = n;
		while(n!=1) {
			n = (n + 7) / 8;
			add(N,8);
			
		}
	
	}


	add(8,4);
	add(4,2);

	add(4,2);
	add(8,2);
	pr();


	
}
int main() {

	int t=1;
	cin>>t;
	while(t--)solve();
	
}