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]. 
from sys import stdin
input = stdin.readline

from math import ceil,sqrt


def help():
	n = int(input().strip())

	toleave = []

	ii = n
	while ii > 2:
		toleave.append(ii)
		ii = ceil(sqrt(ii))

	toleave.sort(reverse = True)

	tt = 0
	toprint = []
	for i in range(3,n):
		if(i not in toleave):
			tt += 1 
			toprint.append([i,n])

	for i in range(len(toleave)-1):
		toprint.append([toleave[i],toleave[i+1]])
		toprint.append([toleave[i],toleave[i+1]])

	# if(len(toleave) == 1):
	# 	toprint.append([toleave[-1],2])
	# 	toprint.append([toleave[-1],2])
	# 	toprint.append([toleave[-1],2])	
	# 	toprint.append([toleave[-1],2])
	# elif(len(toleave) >= 2):
	# 	toprint.append([toleave[-2],2])	
	# 	toprint.append([toleave[-2],2])
	# 	toprint.append([toleave[-2],2])

	# print(len(toprint))
	# for i in toprint:
	# 	print(*i)

	# print(toleave)

	ap = [i for i in range(1,n+1)]
	for i in toprint:
		ap[i[0]-1] = ceil(ap[i[0]-1]/ap[i[1]-1])

	for i in range(2,n):
		while(ap[i] != 1):
			toprint.append([i+1,2])
			ap[i] = ceil(ap[i]/2)

	print(len(toprint))
	for i in toprint:
		print(*i)


# for _ in range(1):
for _ in range(int(input())):
	help()