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 PYTHON3):

Solution:
from sys import stdin
from math import sqrt, floor, ceil


"""
1 2 3 4 5 6 .. 98 99 100
100, 10
100 / 10 = 10

==> 1 2 ... 9 10 1 12 13 ... 98 99 100

==> 1 2 ... 9 10 1 1 13 ... 98 99 100

==> ...

==> 1 2 ... 9 10 1 1 1 ... 1 1 100

==> 1 2 ... 9 10 1 1 1 ... 1 1 10

floor(sqrt(10)) = 3

...

1 2 3 1 .. 1 1 10

==> 

1 2 3 1 .. 1 1 4

floor(sqrt(10)) = 2

==> 

1 2 1 1 .. 1 1 4

==>

1 2 1 1 .. 1 1 2

==>

2 == 2

==> 1 2 1 1 .. 1 1 1
"""

t = int(stdin.readline())

for _ in range(t):
    n = int(stdin.readline())

    A = [x for x in range(n + 1)]
    
    curr = n
    prev = n

    res = []

    while curr > 3:
        curr_sqrt = floor(sqrt(curr))

        for i in range(curr_sqrt + 1, prev):
            res.append((i, n))
            A[i] = ceil(i // n)

        prev = curr_sqrt + 1
        res.append((n, curr_sqrt))
        curr = ceil(curr / curr_sqrt)
    
    if n == 3:
        res.append((3, 2))
        res.append((3, 2))
    else:
        if A[3] == 3:
            while curr > 1:
                res.append((n, 3))
                curr = ceil(curr / 3)
            res.append((3, 2))
            res.append((3, 2))
        else:
            while curr > 1:
                res.append((n, 2))
                curr = ceil(curr / 2)
            res.append((n, 2))

    print(len(res))
    for x, y in res:
        print(x, y)