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 ll;
const int maxn=1e6+7;
const int inf=INT_MAX;
const ll inff=1e18;
const ll mod=1e9+7;
#define pii pair<int,int>
#define mkp make_pair
#define F first
#define S second
#define pb push_back
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(),(v).end()
//#define int ll
 
#ifdef HNO2
#define IOS
#else
#define endl '\n'
#define IOS ios::sync_with_stdio(0); cin.tie(0);
#endif // HNO2
 
int32_t main(){
    IOS
	int t; cin>>t;
	while (t--){
		int n; cin>>n;
		vector<pii> v;
		if (n==3) v.pb({3,2}),v.pb({3,2});
		else if (n<=512){
			for (int i=4;i<n;i++) v.pb({i,i+1});

			int N=n;
			while (N>1) v.pb({n,3}),N=(N+2)/3;
			v.pb({3,2}),v.pb({3,2});
		}
		else{
			for (int i=4;i<512;i++) v.pb({i,i+1});
			for (int i=513;i<n;i++) v.pb({i,i+1});

			int N=n;
			while (N>1) v.pb({n,512}),N=(N+511)/512;
			N=512;
			while (N>1) v.pb({512,3}),N=(N+2)/3;
			v.pb({3,2}),v.pb({3,2});
		}
		assert(sz(v)<=n+5);
		cout<<sz(v)<<endl;
		for (pii i:v) cout<<i.F<<' '<<i.S<<endl;
	}
}