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
testing = len(sys.argv) == 4 and sys.argv[3] == "myTest"
if testing:
    cmd = sys.stdout
    from time import time
    start_time = int(round(time() * 1000)) 
    input = open(sys.argv[1], 'r').readline
    sys.stdout = open(sys.argv[2], 'w')
else:
    input = sys.stdin.readline

from math import ceil, log
# from collections import defaultdict as dd
# from heapq import *
############ ---- Input Functions ---- ############
def intin():
    return(int(input()))
def intlin():
    return(list(map(int,input().split())))
def chrin():
    s = input()
    return(list(s[:len(s) - 1]))
def strin():
    s = input()
    return s[:len(s) - 1]
    
def main():
    n = intin()
    if n == 2:
        print("0")
        return
    if n == 3:
        print("2\n3 2\n3 2")
        return
    if n == 4:
        print("3\n3 4\n4 2\n4 2")
        return
    a = [i+1 for i in xrange(n)]
    x = 15
    ans = []
    for i in xrange(2, n-1):
        if i+1 == x:
            continue
        a[i] = int(ceil(a[i]*1.0/a[i+1]))
        ans.append(str(i+1) +" " +str(i+2))
    if n > x:
        k = int(ceil(log(n, x)))
        while k > 0:
            a[-1] = int(ceil(a[-1]*1.0/x))
            ans.append(str(n)+" "+str(x))
            k -= 1
    k = 4
    while k > 0:
        a[min(n,x)-1] = int(ceil(a[min(n,x)-1] / 2.0))
        ans.append(str(min(n,x))+" 2")
        k -= 1
    # print(a)
    assert sum([1 for y in a if y==1]) == n-1
    assert sum([1 for y in a if y==2]) == 1
    assert len(ans) <= n+5
    print(len(ans))
    print("\n".join(ans))




if __name__ == "__main__":
    for _ in xrange(intin()):
        main()
    # main()

    if testing:
        sys.stdout = cmd
        print(int(round(time() * 1000))  - start_time)