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 <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 400100;

int T, n, a[N];
vector < pair <int, int> > res;

int main() {

	scanf("%d", &T);

	for (int t = 1; t <= T; t++) {
		scanf("%d", &n);
		for (int i = 0; i < n; i++) {
			a[i] = i + 1;
		}
		int d = min(16, n);
		for (int i = 0; i < n; i++) {
			if (a[i] != 1 && a[i] != 2 && a[i] != d && a[i] != n) {
				res.push_back(make_pair(i + 1, n));
			}
		}
		if (n <= 16) {
			while (a[d - 1] > 1) {
				a[d - 1] = (a[d - 1] + 1) / 2;
				res.push_back(make_pair(d, 2));
			}
		} else {
			while (a[n - 1] > 1) {
				res.push_back(make_pair(n, d));
				a[n - 1] = (a[n - 1] + d - 1) / d;
			}
			while (a[d - 1] > 1) {
				res.push_back(make_pair(d, 2));
				a[d - 1] = (a[d - 1] + 1) / 2;
			}
		}
		printf("%d\n", (int) res.size());
		for (int i = 0; i < res.size(); i++) {
			printf("%d %d\n", res[i].first, res[i].second);
		}
		res.clear();
	}

	return 0;

}