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]. 
from collections import Counter
import math

def solve(N):
    nums = [i for i in range(N + 1)]
    cnt = 0
    steps = []

    can = set()
    n = N
    while n > 2:
        can.add(n)
        n = math.ceil(math.sqrt(n))

    for i in range(3, N, 1):
        if i in can:
            continue
        nums[i] = math.ceil(nums[i] / nums[-1])
        cnt += 1
        steps.append((i, N))

    for idx in sorted(can, reverse=True):
        target = math.ceil(math.sqrt(nums[idx]))
        while nums[idx] > 1:
            nums[idx] = math.ceil(nums[idx] / nums[target])
            steps.append((idx, target))

    cn = Counter(nums)
    assert len(cn) == 3
    assert cn[0] == 1
    assert cn[1] == N - 1
    assert cn[2] == 1
    assert cnt <= N + 5

    return steps

T = int(input())
for _ in range(T):
    N = int(input())
    ans = solve(N)
    print(len(ans))
    for x, y in ans:
        print(x, y)
