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;
#define ll          long long int
#define pb          push_back
#define pf          push_front
#define ppb         pop_back
#define ppf         pop_front
#define TC          ll t;cin>>t; for(ll _ = 1;_ <=t;_ ++)
#define bs          binary_search
#define pll         pair<ll,ll>
#define ppll        pair<ll,pll>
#define vvll        vector<vector<ll>>
#define sll         stack<ll>
#define qll         queue<ll>
#define pqll        priority_queue<ll>
#define pqllmin     priority_queue<ll, vector<ll>, greater<ll>>
#define vll         vector<ll>
#define vpll        vector<pll>
#define f           first
#define s           second
#define MOD         1000000007
#define mxn         10e18
#define ln          "\n"
#define print       cout<<"Case #"<<_<<": "<<ans<<ln;
#define sps(x)      fixed<<setprecision(x)
#define popcount(x) __builtin_popcountll(x)
#define mp			make_pair

template <typename T> istream& operator >> (istream &in , vector<T> &v) { for (auto &i : v) cin >> i; return in; }
template <typename T> istream& operator >> (istream &in , pair<T , T> &p) { cin >> p.f >> p.s; return in; }
template <typename T> ostream& operator << (ostream &out , vector<T> &v) { for (auto &i : v) cout << i << " "; return out; }
template <typename T> ostream& operator << (ostream &out , pair<T , T> &p) { cout << p.f << " " << p.s << " "; return out; }
template <typename T> istream& operator >> (istream &in , vector<pair<T , T>> &v) { for (auto &i : v) cin >> i.f >> i.s; return in; }
template <typename T> ostream& operator << (ostream &out , vector<pair<T , T>> &v) { for (auto &i : v) cout << i.f << " " << i.s << ln; return out; }



void solve()
{
	TC

	{
		ll n;
		cin >> n;
		vpll ans;
		//ll x = n;
		//ll x = n;
		//	ll a = n;

		ll abc[6];
		abc[0] = 2;
		for (ll i =  1; i < 6 ; i++)
			abc[i] = abc[i - 1] * abc[i - 1];

		ll a = abc[0] , b ;
		ll indx = 0;
		for (ll i = 0 ; i < 6 ; i++)
		{
			if (n <= abc[i])
			{
				a = abc[i - 1];
				indx = i - 1;
				b = n;
				break;
			}

		}

		if (n <= 2)
			cout << 0 << ln;
		else
		{
			ll num = n - 1;
			//	cout << ln << a << " ";
			while (num >= 2)
			{
				while (num > a)
				{
					ans.pb(mp(num , b));
					num--;
				}
				ans.pb(mp(b , a));
				ans.pb(mp(b , a));
				indx--;
				b = a;
				a = abc[indx];
				num--;

			}
			cout << ans.size() << ln;
			for (ll i  = 0  ; i < ans.size() ; i++)
			{
				cout << ans[i].first << " " << ans[i].second << ln;
			}
		}




	}
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	solve();
	return 0;
}