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]. 
import sys
range = xrange
input = raw_input

def f(a,b):
    a = (a + b - 1) // b
    a = (a + b - 1) // b
    return a

def g(a,b):
    return (a + b - 1) // b

inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
out = []

t = inp[ii]; ii += 1
for _ in range(t):
    n = inp[ii]; ii += 1
    A = [i for i in range(n + 1)]

    tmp = []
    j = n
    while n > 2:
        while j > 2 and f(n, j - 1) == 1:
            j -= 1
        
        for i in range(j + 1, n):
            A[i] = g(A[i], A[n])
            tmp.append('%d %d' % (i, n))
        
        A[n] = g(A[n], A[j])
        A[n] = g(A[n], A[j])
        tmp.append('%d %d' % (n,j))
        tmp.append('%d %d' % (n,j))
        n = j

    out.append(str(len(tmp)))
    out += tmp
    #print A[1:]

print '\n'.join(out)
