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:
def solve1(n):

    a = []
    c = 0

    for i in range(3, n):
        a.append(i)
        a.append(i + 1)
        c = c + 1
    t = 0
    z = n
    while z != 1:
        z = (z + 1) // 2
        t = t + 1

    for i in range(t):
        a.append(n)
        a.append(2)
    print(c + t)
    d = 0
    for i in range(0, len(a)):
        print(a[i], end=" ")
        d = d + 1
        if d == 2:
            d = 0
            print()




def solve2():
    n = int(input())
    if n<=8:
        solve1(n)
    else:
        a = []
        c = 0

        for i in range(3, n):
            if i != 8:
                a.append(i)
                a.append(n)

        nn = n
        while nn != 1:
            a.append(n)
            a.append(8)
            nn = (nn + 7) // 8

        a.append(8)
        a.append(2)
        a.append(8)
        a.append(2)
        a.append(8)
        a.append(2)
        print(len(a) // 2)
        d = 0
        for i in range(0, len(a)):
            print(a[i], end=" ")
            d = d + 1
            if d == 2:
                d = 0
                print()

for testis in range(int(input())):
    solve2()