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]. 
t=int(input())
from math import *

#t=1
def f(n,arr,val,i):
    if val==0:
        return True
    if n==0 or val<0:
        return False
    if (arr[i]>val):
        return f(n-1,arr,val,i+1)
    return f(n-1,arr,val,i+1) or f(n-1,arr,val-arr[i],i+1)

for i in range(t):   
    n=int(input())
    count=0
    arr=[]
    
    if (n>10):
        for i in range(n-1,2,-1):
            if i!=10 and i!=4:
                #print(i,i+1)
                #print(2)
                arr.append([str(i),str(n)])
                count+=1
        c=n;
        while c!=1:
            #print(n,10)
            #print(22)
            arr.append([str(n),str(10)])
            c=ceil(c/10)
            count+=1
        c=10
        while c!=1:
            #print(10,4)
            #print(23)
            arr.append([str(10),str(4)])
            c=ceil(c/4)
            count+=1
        c=4
        while c!=1:
            #print(4,2)
            arr.append([str(4),str(2)])
            c=c//2
            count+=1
        print(count)
        for jp in range(count):
            print(' '.join(arr[jp]))
            els=1
    else:
        for i in range(n-1,2,-1):
            #print(i,i+1)
            arr.append([str(i),str(n)])
            count+=1
        c=n;
        while c!=1:
            #print(n,10)
            arr.append([str(n),str(2)])
            c=ceil(c/2)
            count+=1
        print(count)
        for jp in range(count):
            print(' '.join(arr[jp]))
            els=1;
        
        
        
        
            
        
    
    
