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:
    /*author:	Slns*/

#include <bits/stdc++.h>

#define _USE_MATH_DEFINES
#define pb push_back
#define all(v) v.begin(),v.end()
#define ff first
#define ss second

using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair < int, int > pii;
typedef vector<int> vi;
typedef vector<vector<int> > vvi;

void NITRO(){
	//OneInOneOut
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
}

const int N = 1e+5 * 3 + 11;
const ll MOD = 1e+9 + 7; // 888244353;
const int inf = int(1e9);
const ll INF = ll(1e16);

void SOLVE(/*WATLE*/) {
	int n;
	cin >> n;
	vector <pll> v;
	while(n > 2) {
		int t = sqrt(n);
		if(t * t != n)t++;
		for(int i = t + 1; i < n; i++) v.pb({i, n});
		v.pb({n, t}), v.pb({n, t});
		n = t;
	}
	cout << v.size() << '\n';
	for(int i = 0; i < int(v.size()); i++){
		cout << v[i].ff << " " << v[i].ss << '\n';
	}
}

int main() {
    //  freopen("slalom.in", "r", stdin);
    //  freopen("slalom.out", "w", stdout);
    NITRO();
    int times = 1;
	cin >> times;
    for(int i = 1; i <= times; i++) {
        //cout << "Case #" << t << ": ";
        SOLVE();
    }
    return 0;
}