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>
using namespace std;
typedef long long int ll;
typedef long double ld;
#define endl '\n';

void solve(){
	ll n;
	cin >> n;
	vector <pair<ll,ll>> ans;
	if(n<=32){
		for(ll i=3;i<n;i++){
			ans.push_back(make_pair(i,n));
		}
		ll k = n;
		while(k>1){
			ans.push_back(make_pair(n,2));
			k = (k+1)/2;
		}
	}
	else{
		for(ll i=3;i<n;i++){
			if(i!=32){
				ans.push_back(make_pair(i,n));
			}
		}
		ll k = n;
		while(k>1){
			ans.push_back(make_pair(n,32));
			k = (k+31)/32;
		}
		k = 32;
		while(k>1){
			ans.push_back(make_pair(32,2));
			k = (k+1)/2;
		}
	}
	cout << (ll)ans.size() << endl;
	for(auto x : ans){
		cout << x.first << " " << x.second << endl;
	}
}

int main(){
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int tc;
	tc = 1;
	cin >> tc;
	for(int i=1;i<=tc;i++){
		solve();
	}
	return 0;
}