What Is It?
Lunar rover finally reached planet X. After landing, he met an obstacle, that contains permutation $$$p$$$ of length $$$n$$$. Scientists found out, that to overcome an obstacle, the robot should make $$$p$$$ an identity permutation (make $$$p_i = i$$$ for all $$$i$$$).Unfortunately, scientists can't control the robot. Thus the only way to make $$$p$$$ an identity permutation is applying the following operation to $$$p$$$ multiple times:   Select two indices $$$i$$$ and $$$j$$$ ($$$i \neq j$$$), such that $$$p_j = i$$$ and swap the values of $$$p_i$$$ and $$$p_j$$$. It takes robot $$$(j - i)^2$$$ seconds to do this operation.  Positions $$$i$$$ and $$$j$$$ are selected by the robot (scientists can't control it). He will apply this operation while $$$p$$$ isn't an identity permutation. We can show that the robot will make no more than $$$n$$$ operations regardless of the choice of $$$i$$$ and $$$j$$$ on each operation.Scientists asked you to find out the maximum possible time it will take the robot to finish making $$$p$$$ an identity permutation (i. e. worst-case scenario), so they can decide whether they should construct a new lunar rover or just rest and wait. They won't believe you without proof, so you should build an example of $$$p$$$ and robot's operations that maximizes the answer.For a better understanding of the statement, read the sample description.
For $$$n = 2$$$, $$$p$$$ can be either $$$[1, 2]$$$ or $$$[2, 1]$$$. In the first case $$$p$$$ is already identity, otherwise robot will make it an identity permutation in $$$1$$$ second regardless of choise $$$i$$$ and $$$j$$$ on the first operation.For $$$n = 3$$$, $$$p$$$ can be equals $$$[2, 3, 1]$$$.  If robot will select $$$i = 3, j = 2$$$ on the first operation, $$$p$$$ will become $$$[2, 1, 3]$$$ in one second. Now robot can select only $$$i = 1, j = 2$$$ or $$$i = 2, j = 1$$$. In both cases, $$$p$$$ will become identity in one more second ($$$2$$$ seconds in total). If robot will select $$$i = 1, j = 3$$$ on the first operation, $$$p$$$ will become $$$[1, 3, 2]$$$ in four seconds. Regardless of choise of $$$i$$$ and $$$j$$$ on the second operation, $$$p$$$ will become identity in five seconds. We can show, that for permutation of length $$$3$$$ robot will always finish all operation in no more than $$$5$$$ seconds.
The first line of input contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases.Each of next $$$t$$$ lines contains the single integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) – the length of $$$p$$$.Note, that $$$p$$$ is not given to you. You should find the maximum possible time over all permutations of length $$$n$$$.It is guaranteed, that the total sum of $$$n$$$ over all test cases doesn't exceed $$$10^5$$$.
For each test case in the first line, print how many seconds will the robot spend in the worst case.In the next line, print the initial value of $$$p$$$ that you used to construct an answer.In the next line, print the number of operations $$$m \leq n$$$ that the robot makes in your example.In the each of next $$$m$$$ lines print two integers $$$i$$$ and $$$j$$$ — indices of positions that the robot will swap on this operation. Note that $$$p_j = i$$$ must holds (at the time of operation).
for tests in range(int(input())):
    n = int(input())
    AA = [(1, i) for i in range((n + 2) // 2, n)] + [(i, n) for i in range((n + 2) // 2 - 1, 0, -1)]
    L = list(range(1, n + 1))
    sc = 0
    for (x, y) in AA[::-1]:
        (L[x - 1], L[y - 1]) = (L[y - 1], L[x - 1])
        sc += (x - y) ** 2
    print(sc)
    print(*L)
    print(len(AA))
    for (x, y) in AA:
        print(y, x) if L[x - 1] == y else print(x, y)
        (L[x - 1], L[y - 1]) = (L[y - 1], L[x - 1])
import sys
input = sys.stdin.readline
t = int(input())
for tests in range(t):
    n = int(input())
    AA = []
    for i in range((n + 2) // 2, n):
        AA.append((1, i))
    for i in range((n + 2) // 2 - 1, 0, -1):
        AA.append((i, n))
    L = list(range(1, n + 1))
    sc = 0
    for (x, y) in AA[::-1]:
        (L[x - 1], L[y - 1]) = (L[y - 1], L[x - 1])
        sc += (x - y) ** 2
    print(sc)
    print(*L)
    print(len(AA))
    for (x, y) in AA:
        if L[x - 1] == y:
            print(y, x)
        else:
            print(x, y)
        (L[x - 1], L[y - 1]) = (L[y - 1], L[x - 1])
for tests in range(int(input())):
    n = int(input())
    AA = [(1, i) for i in range((n + 2) // 2, n)] + [(i, n) for i in range((n + 2) // 2 - 1, 0, -1)]
    L = list(range(1, n + 1))
    sc = 0
    for (x, y) in AA[::-1]:
        (L[x - 1], L[y - 1]) = (L[y - 1], L[x - 1])
        sc += (x - y) ** 2
    print(sc)
    print(*L)
    print(len(AA))
    for (x, y) in AA:
        print(y, x) if L[x - 1] == y else print(x, y)
        (L[x - 1], L[y - 1]) = (L[y - 1], L[x - 1])