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]. 

def readInt():
    return int(input())

def readLine():
    return [int(s) for s in input().split(" ")]

def readString():
    return input()

def isGoodPivot(k,n):
    s1 = k
    s2 = n
    S1 = s1>s2
    while True:
        if S1:
            s1 = -(s1 // -s2)
        else:
            s2 = -(s2 // -s1)

        S1 = not S1

        if(min(s1,s2) == 1):
            if(max(s1,s2)==2):
                return True
            else:
                return False

from math import sqrt,ceil
def calPath(n):
    y = lambda x: ceil(sqrt(x))
    ret = []
    while(n > 2):
        n = y(n)
        ret.append(n)

    # print(ret)
    return ret


def ask(n):
    path = calPath(n)
    stuff = set(path)

    # pivot = choosePivot(n)
    ret = []

    for i in range(2,n,1):
        if i in stuff:
            continue
        ret.append((i,n))

    last = n
    for t in path:
        ret.append((last,t))
        ret.append((last, t))
        last = t
    # s1 = n
    # s2 = pivot

    # S1 = True
    # while min(s1,s2) != 1 or max(s1,s2) != 2:
    #     # if(s1 > s2):
    #     #     S1 = True
    #     # else:
    #     #     S1 = False
    #
    #     if S1:
    #         ret.append((n,pivot))
    #         S1 = not S1
    #         s1 = -(s1 // -s2)
    #     else:
    #         ret.append((pivot, n))
    #         S1 = not S1
    #         s2 = -(s2 // -s1)
    #
    #
    # # print("Len" len(ret))
    return ret



t = readInt()
for _ in range(t):
    n = readInt()
    result = ask(n)
    print(len(result))
    for a,b in result:
        print(a,b)

    # print(len(result))