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 REP(i,a,n) for(int i=(a);i<(n);++i)
#define BREP(i,a,n) for(int i=(a);i>(n);--i)
#define TRAV(i, v) for(auto &i: v)
#define fi first
#define se second
#define pb push_back
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(), (x).end()
#define INDEX(arr,ind)	(upper_bound(ALL(arr),ind)-arr.begin())
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define pow10(n) stoll("1"+string(n,'0'))
#define EPS 1e-9
// #define LOCAL 1

typedef long long ll; 
typedef long double ld;
typedef pair<int, int> ii; 
typedef pair<ll, ll> pll; 
typedef vector < int > vi;
typedef vector < ll > vll;
typedef vector<ii> vii; 
typedef map<int, int> mii;

const long long INF = 1LL << 60;

vii v;
void rec(int x){
	if(x<=2) return;
	int mid=ceil(sqrt(x));
	for (int i = mid+1; i < x; ++i)
	{
		v.pb({i,x});
	}
	REP(i,0,2) v.pb({x,mid});
	rec(mid);
}

void solve(){
	v.clear();
	int n; cin >> n;
	rec(n);
	cout << SZ(v);
	REP(i,0,SZ(v)) cout << endl << v[i].fi << " " << v[i].se;
	cout << endl;	
}

int main(){
	
	#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("error.txt", "w", stderr);		
	#endif

	int T;
	cin >> T;
	for (int tc = 1; tc <= T; ++tc)
	{
		// cout << "Case #" << tc << ": ";
		solve();
	}
	return 0;
}