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]. 
# Hello World program in Python

def readLine():
    return [int(s) for s in input().split(" ")]

zbior = {2,4,16,256,65536}

def solve():
    n = int(input())
    ans = []
    
    maxi = 2
    
    for i in range(2,n + 1):
        y = {i}
        if(y.issubset(zbior)):
            continue
        else:
            if(i < n):
                ans.append([i , n])
            else:
                while(maxi * maxi <= n):
                    maxi = maxi * maxi
                ans.append([n, maxi])
                ans.append([n, maxi])
    
    while(maxi * maxi <= n):
        maxi = maxi * maxi
    
    specjalne = []
    cur = 2
    while(cur <= maxi):
        specjalne.append(cur);
        cur = cur * cur
    
    specjalne.reverse()
    
    p = 0
    
    for cur in specjalne:
        if(cur == 2):
            break
        pop = specjalne[p + 1]
        akt = specjalne[p]
        
        ans.append([akt, pop])
        ans.append([akt, pop])
        
        p = p + 1
    
    print(len(ans))
    
    for x in ans:
        print(x[0], x[1])
    
    
    
tt = int(input())
for x in range(tt):
    solve()

