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())
lg = [0] * (2 * 10 ** 5 + 5)
for i in range(2, 2 * 10 ** 5 + 1):
    lg[i] = lg[i // 2] + 1
for _ in range(t):
    n = int(input())
    if n == 3:
        print(2)
        print(3, 2)
        print(3, 2)
    else:
        max_lg = lg[n]
        keep = []
        cand = set(range(2, n + 1))
        while max_lg > 0:
            keep.append(1 << max_lg)
            cand.remove(1 << max_lg)
            max_lg //= 2
        op = []
        keep.sort(reverse=True)
        if 1 << lg[n] != n:
            cand.remove(n)
            cand.add(2)
        for elm in cand:
            op.append((elm, n))
        if 1 << lg[n] != n:
            op.append((n, 1 << lg[n]))
        for i in range(1, len(keep)):
            cnt = lg[keep[i - 1]] // lg[keep[i]]
            if lg[keep[i - 1]] % lg[keep[i]] != 0:
                cnt += 1
            if i == len(keep) - 1 and (1 << lg[n]) != n:
                for _ in range(cnt):
                    op.append((keep[i - 1], n))
                continue
            for _ in range(cnt):
                op.append((keep[i - 1], keep[i]))
        print(len(op))
        for p in op:
            print(*p)
