#include <bits/stdc++.h>
using namespace std;

vector<pair<int, int>> V;

int f(int x, int y) {
	return (x+y-1)/y;
}

bool policz(int x, int y) {
	int ruchy = 0;
	while((x!=2 or y!=1) and ruchy < 8) {
		int tmp = f(x, y);
		x = tmp;
		if(x < y)
			swap(x, y);
		ruchy++;
	}
	if(x == 2 and y == 1) 
		return true;
	return false;
}

void wypisz(int x, int y) {
	int ilosc = 0;
	for(int i=2;i<x;i++) {
		if(i != y) {
			V.push_back({i, x});
			ilosc++;
		}
	}
	int xx = x;
	int yy = y;
	while(x!=2 or y!=1) {
		int tmp = f(max(x, y), min(x, y));
		if(x > y) {
			x = tmp;
			ilosc++;
			V.push_back({xx, yy});
		} else {
			y = tmp;
			ilosc++;
			V.push_back({yy, xx});
		}
	}
	cout << ilosc << "\n";
	for(auto para:V) 
		cout << para.first << " " << para.second << "\n";
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	int t;
	cin >> t;
	while(t--) {
		V.clear();
		int n;
		cin >> n;
		for(int i=1;i<n;i++) {
			if(policz(n, i)) {
				wypisz(n, i);
				break;
			}
		}
	}
}

### Prompt

Please formulate a CPP solution to the following 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]. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

vector<pair<int, int>> V;

int f(int x, int y) {
	return (x+y-1)/y;
}

bool policz(int x, int y) {
	int ruchy = 0;
	while((x!=2 or y!=1) and ruchy < 8) {
		int tmp = f(x, y);
		x = tmp;
		if(x < y)
			swap(x, y);
		ruchy++;
	}
	if(x == 2 and y == 1) 
		return true;
	return false;
}

void wypisz(int x, int y) {
	int ilosc = 0;
	for(int i=2;i<x;i++) {
		if(i != y) {
			V.push_back({i, x});
			ilosc++;
		}
	}
	int xx = x;
	int yy = y;
	while(x!=2 or y!=1) {
		int tmp = f(max(x, y), min(x, y));
		if(x > y) {
			x = tmp;
			ilosc++;
			V.push_back({xx, yy});
		} else {
			y = tmp;
			ilosc++;
			V.push_back({yy, xx});
		}
	}
	cout << ilosc << "\n";
	for(auto para:V) 
		cout << para.first << " " << para.second << "\n";
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	int t;
	cin >> t;
	while(t--) {
		V.clear();
		int n;
		cin >> n;
		for(int i=1;i<n;i++) {
			if(policz(n, i)) {
				wypisz(n, i);
				break;
			}
		}
	}
}

```