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]. 
#include<bits/stdc++.h>
using namespace std;

#define double long double
#define int long long
#define ALL(x) x.begin(),x.end()
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
#define pb(x) push_back(x)
#define eb emplace_back
#define SORT(ls) sort(ls.begin(), ls.end())
#define DESC(ls) sort(ls.rbegin(), ls.rend()) 
#define mp(x, y) make_pair(x, y)
#define INF 1000000000
#define MOD 998244353 
#define F(i, a, n) for(i=a;i<n;i++)
#define SI(x) cin>>x
#define DI(x, y) cin>>x>>y
#define TI(x, y, z) cin>>x>>y>>z
#define FI(w, x, y, z) cin>>w>>x>>y>>z
#define SO(x) cout<<x<<"\n"
#define DO(x, y) cout<<x<<" "<<y<<"\n"
#define TO(x, y, z) cout<<x<<" "<<y<<" "<<z<< "\n"
#define FO(w, x, y, z) cout<<w<<" "<<x<<" "<<y<<" "<<z<<"\n"
#define FASTER ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)

int c(int a, int b)
{
	return a/b + ((a%b) > 0);
}

int32_t main(void)
{
    FASTER;
    int i, j;
	int t;
	SI(t);
	while(t--)
	{
		
		int n;
		SI(n);
		int a[n + 1];
		F(i, 1, n + 1)
			a[i] = i;
		vii ans;
		F(i, 3, n)
		{
			if(i == 8)
				continue;
			ans.pb(mp(i, n));
			a[i] = c(a[i], a[n]);
		}
		
		int rem = (n > 8) ? 8 : 2;
		while(a[n] > 1)
		{
			ans.pb(mp(n, rem));
			a[n] = c(a[n], rem);
		}	
		if(n > 8)
		{
			a[8] = 1;
			ans.pb(mp(8, 2));
			ans.pb(mp(8, 2));
			ans.pb(mp(8, 2));
		}
		int res[3];
		res[2] = res[1] = 0;
		F(i, 1, n + 1)
		{
			assert(a[i] < 3);
			res[a[i]]++;
		}
		assert(res[a[2]] == 1);
		assert(res[a[1]] == n - 1);
		assert(ans.size() <= n + 5);
		SO(ans.size());
		for(auto i: ans)
			DO(i.first, i.second);
	}
	
	    
    return 0;
}