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]. 
#!/usr/bin/env pypy3

import math

def ans(N):

	arr = list(range(1,N+1))

	ret = []
	if N <= 64:
		for i in range(3, N):
			ret += [(i, N)]
			arr[i-1] = math.ceil(arr[i-1] / N)
		while arr[-1] != 1:
			ret += [(N, 2)]
			arr[-1] = math.ceil(arr[-1] / 2)

	else:
		REDUCER = 13

		for i in range(3, N):
			if i == REDUCER: continue

			ret += [(i, N)]
			arr[i-1] = math.ceil(arr[i-1] / N)

		while arr[-1] != 1:
			# print("reducing max")
			ret += [(N, REDUCER)]
			arr[-1] = math.ceil(arr[-1] / REDUCER)

		while arr[REDUCER-1] != 1:
			# print(f"reducing {REDUCER}")
			ret += [(REDUCER, 2)]
			arr[REDUCER-1] = math.ceil(arr[REDUCER-1] / 2)

	assert(len(ret) <= N+5)
	assert(arr.count(1) == N-1)
	assert(arr.count(2) == 1)

	print(len(ret))
	for r in ret:
		print(*r)

for _ in range(int(input())):
	N = int(input())
	ans(N)
