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
input = sys.stdin.readline
import math
import copy
import collections
from collections import deque
inf = 10**18

for _ in range(int(input())):
    n = int(input())
    ans = []
    special = []
    num = n
    exp = 1
    while num>2:
        special.append(num)
        num = math.ceil(num/4**exp)
        exp+=1
    special.append(2)
    # print(special)
    a = []
    for i in range(len(special)-1):
        a.append(math.ceil(special[i]/special[i+1]))
    a.append(2)
    s = set(special)
    for i in range(1,n):
        if i!=1 and i not in special:
            ans.append([i,i+1])
    index = dict()
    for i in range(len(special)):
        index[i] = special[i]
    for i in range(len(special)-1):
        ans.append([special[i],special[i+1]])
    # print(a)
    for i in range(len(a)-2,0,-1):
        a[i] = math.ceil(a[i]/a[i-1])
        ans.append([index[i],index[i-1]])
    # print(a)
    for i in range(len(a)-2,0,-1):
        if a[i]!=1 and a[i-1]!=1:
            a[i] = math.ceil(a[i]/a[i-1])
            ans.append([index[i],index[i-1]])
    # print(a)
    while a[0]>1:
        a[0] = math.ceil(a[0]/2)
        ans.append([index[0],index[len(a)-1]])
    # print(a)
    print(len(ans))
    for i in range(len(ans)):
        u,v = ans[i]
        print(u,v)