Problem: 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 ≠ 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.

Input

The first line of input contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

Each of next t lines contains the single integer n (2 ≤ n ≤ 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.

Output

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 ≤ 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).

Example

Input


3
2
3
3


Output


1
2 1
1
2 1
5
2 3 1
2
1 3
3 2
5
2 3 1
2
1 3
2 3

Note

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.

Solution (in PYTHON3):

Solution:
import sys
readline = sys.stdin.readline

def parorder(Edge, p):
    N = len(Edge)
    par = [0]*N
    par[p] = -1
    stack = [p]
    order = []
    visited = set([p])
    ast = stack.append
    apo = order.append
    while stack:
        vn = stack.pop()
        apo(vn)
        for vf in Edge[vn]:
            if vf in visited:
                continue
            visited.add(vf)
            par[vf] = vn
            ast(vf)
    return par, order
 
def getcld(p):
    res = [[] for _ in range(len(p))]
    for i, v in enumerate(p[1:], 1):
        res[v].append(i)
    return res
 
 




T = int(readline())
Ans = []
INF = 2*10**9+7
for qu in range(T):
    N = int(readline())
    Edge = [[] for _ in range(N)]
    m = N-1
    Edge[0].append(m)
    Edge[m].append(0)
    ans = m*m
    for i in range(1, N-1):
        if i-0 > m-i:
            Edge[i].append(0)
            Edge[0].append(i)
            ans += i*i
        else:
            Edge[i].append(m)
            Edge[m].append(i)
            ans += (m-i)**2
    Ans.append(str(ans))
    P, L = parorder(Edge, 0)
    pe = list(range(1, N+1))
    res = []
    for l in L[:0:-1]:
        p = P[l]
        res.append((l+1, p+1))
    for a, b in res[::-1]:
        a -= 1
        b -= 1
        pe[a], pe[b] = pe[b], pe[a]
    Ans.append(' '.join(map(str, pe)))
    Ans.append(str(N-1))
    
    Ans.append('\n'.join(['{} {}'.format(*a) for a in res]))
    
        
    
    
print('\n'.join(Ans))