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:
import sys
import math
def II():
	return int(sys.stdin.readline())
 
def LI():
	return list(map(int, sys.stdin.readline().split()))
 
def MI():
	return map(int, sys.stdin.readline().split())
 
def SI():
	return sys.stdin.readline().strip()

def FACT(n, mod):
    s = 1
    facts = [1]
    for i in range(1,n+1):
        s*=i
        s%=mod
        facts.append(s)
    return facts[n]

def C(n, k, mod):
    return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod

t = II()
for q in range(t):
    n = II()
    d = []
    temp = n
    xx = 8
    while xx<temp:
        d.append(xx)
        xx*=8
    d.append(temp)
    d.reverse()
    ans = []
    for i in range(3,n):
        if i not in d:
            ans.append([i, i+1])
    for i in range(len(d)-1):
        ans.append([d[i], d[i+1]])
    for i in range(len(d)-1):
        ans.append([d[i], d[i+1]])
    temp = d[-1]
    while temp!=1:
        ans.append([d[-1], 2])
        temp = math.ceil(temp/2)
    print(len(ans))
    for i in ans:
        print(*i)