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 io
import os

from collections import Counter, defaultdict, deque


def solve(N,):
    ans = []

    A = list(range(1, N + 1))

    def ceilDiv(i, j):
        ans.append(str(i + 1) + " " + str(j + 1))
        assert i != j
        x = A[i]
        y = A[j]
        A[i] = (x + y - 1) // y

    for i in reversed(range(1, N - 1)):
        if (A[N - 1] + A[i] - 1) // A[i] >= A[i]:
            ceilDiv(N - 1, i)
        if A[i] != 2:
            ceilDiv(i, N - 1)
    assert A[1] == 2
    while A[N - 1] != 1:
        ceilDiv(N - 1, 1)
    '''
    assert len(ans) <= N + 5
    if Counter(A) != Counter({1: N - 1, 2: 1}):
        assert False
    '''
    return str(len(ans)) + "\n" + "\n".join(ans)


# for n in range(3, 2 * 10 ** 5):
# solve(n)
# solve(2 * 10 ** 5)


if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    TC = int(input())
    for tc in range(1, TC + 1):
        (N,) = [int(x) for x in input().split()]
        ans = solve(N,)
        print(ans)
