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 endl '\n'
#define fastIO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define int long long int
#define pb push_back
#define ff first
#define ss second
#define all(v) (v).begin(), (v).end()
#define mod (int)(998244353)
#define PI 3.14159265358979323846264338327950L
// ------------------ Debugging ------------------
#define trace(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args);cerr << endl; }
void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cerr << "[ " << *it << " = " << a << " ] ";
    err(++it, args...);
}
template <typename Tk, typename Tv> ostream& operator<<(ostream& os, const pair<Tk, Tv> &p){os << "{" << p.first << ',' << p.second << "}";return os;}
template <typename T> ostream& operator<<(ostream& os, const vector<T> &p){os << "[ "; for (T x: p) os << x << " "; os << "]" << endl; return os;}
template <typename T> ostream& operator<<(ostream& os, const set<T> &p){os << "{ "; for (T x: p) os << x << " "; os << "}" << endl; return os;}
template <typename Tk, typename Tv> ostream& operator<<(ostream& os, const map<Tk, Tv> &p){os << "{ "; for (pair<Tk, Tv> x: p) os << x << " "; os << "}" << endl; return os;}

#define imie(args) cerr << #args << ": "; cerr << args;
// -----------------------------------------------

void solve(int n, vector<pair<int, int>>&ans) {
	if (n == 2)
		return;

	int y = max(1ll, (int)sqrt(n)-1);
	while (y < (n+y-1)/y) {
		y++;
	}
	for (int pos = y+1; pos < n; pos++) {
		ans.emplace_back(pos, n);
	}
	ans.emplace_back(n, y);
	ans.emplace_back(n, y);
	solve(y, ans);
}

signed main(){

    fastIO

    int t;
    cin >> t;
    while (t--) {
    	int n;
    	cin >> n;
    	vector<pair<int, int>> ans;
    	solve(n, ans);
    	cout << (int) ans.size() << "\n";
    	for (auto&it : ans) {
    		cout << it.ff << ' ' << it.ss << "\n";
    	}
    }
    
    return 0;
}