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 math import  ceil
def ceiling(n):
    blanck=[]
    lst=[i for i in range(1,n+1)]
    for i in range(2,len(lst)-1):
        lst[i]=1
        blanck.append([i+1,len(lst)])
    maxi=max(lst)
    while maxi!=1:
        maxi=ceil(maxi/2)
        blanck.append([len(lst),2])
    if len(blanck)<=n+5:
        print(len(blanck))
        for i in blanck:
            print(*i)
        return ""
    else:
        cnst=25
        blanck = []
        lst = [i for i in range(1, n + 1)]
        for i in range(2, len(lst) - 1):
            if lst[i]!=cnst:
                lst[i] = 1
                blanck.append([i + 1, len(lst)])

        maxi = max(lst)
        while maxi != 1:
            maxi = ceil(maxi / cnst)
            blanck.append([len(lst), cnst])
        lst[-1]=1
        maxi=max(lst)
        while maxi != 1:
            maxi = ceil(maxi / 2)
            blanck.append([cnst,2])
        if len(blanck) <= n + 5:
            print(len(blanck))
            for i in blanck:
                print(*i)
            return ""
for i in range(int(input())):
    print(ceiling(int(input())))
