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>
 
#define endl '\n'
#define ll long long int
#define let(x, a) __typeof(a) x(a)
#define all(a) (a).begin(), (a).end()
#define present(c, x) ((c).find(x) != (c).end())
#define tr(v, it) for (let(it, v.begin()); it != v.end(); it++)
#define rtr(v, it) for (let(it, v.rbegin()); it != v.rend(); it++)
 
using namespace std;
 
#ifdef DEBUG
#define trace1(x)					cerr << #x << ": " << x << endl;
#define trace2(x, y)				cerr << #x << ": " << x << " | " << #y << ": " << y << endl;
#define trace3(x, y, z)				cerr << #x << ": " << x << " | " << #y << ": " << y << " | " << #z << ": " << z << endl;
#define trace4(a, b, c, d)			cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << endl;
#define trace5(a, b, c, d, e)		cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << endl;
#define trace6(a, b, c, d, e, f)	cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << " | " << #f << ": " << f << endl;
#else
#define trace1(x)
#define trace2(x, y)
#define trace3(x, y, z)
#define trace4(a, b, c, d)
#define trace5(a, b, c, d, e)
#define trace6(a, b, c, d, e, f)
#endif

int cubert(int n) {
	for(int i=1;;i++) {
		if(i*i*i>n) return i-1;
	}
	return 0;
}

int main() {
	int t,n,m,c;
	cin>>t;
	while(t--) {
		cin>>n;
		m=cubert(n);
		vector<pair<int,int>> v;
		for(int i=3;i<m;i++) {
			v.push_back({i,m});
		}
		for(int i=max(3,m+1);i<n;i++) {
			v.push_back({i,n});
		}
		v.push_back({n,m});
		v.push_back({n,m});
		if(m!=2) v.push_back({m,n});
		c=n/m+(n%m ? 1 : 0);
		c=c/m+(c%m ? 1 : 0);
		while(c!=1) {
			v.push_back({n,2});
			c=c/2+(c%2 ? 1 : 0);
		}
		cout<<v.size()<<endl;
		assert(v.size()<=n+5);
		for(auto p:v) cout<<p.first<<" "<<p.second<<endl;
	}
	return 0;
}