Weird LCM Operations
Given an integer $$$n$$$, you construct an array $$$a$$$ of $$$n$$$ integers, where $$$a_i = i$$$ for all integers $$$i$$$ in the range $$$[1, n]$$$. An operation on this array is defined as follows:  Select three distinct indices $$$i$$$, $$$j$$$, and $$$k$$$ from the array, and let $$$x = a_i$$$, $$$y = a_j$$$, and $$$z = a_k$$$.  Update the array as follows: $$$a_i = \operatorname{lcm}(y, z)$$$, $$$a_j = \operatorname{lcm}(x, z)$$$, and $$$a_k = \operatorname{lcm}(x, y)$$$, where $$$\operatorname{lcm}$$$ represents the least common multiple.  Your task is to provide a possible sequence of operations, containing at most $$$\lfloor \frac{n}{6} \rfloor + 5$$$ operations such that after executing these operations, if you create a set containing the greatest common divisors (GCDs) of all subsequences with a size greater than $$$1$$$, then all numbers from $$$1$$$ to $$$n$$$ should be present in this set.After all the operations $$$a_i \le 10^{18}$$$ should hold for all $$$1 \le i \le n$$$.We can show that an answer always exists.
In the third test case, $$$a = [1, 2, 3, 4, 5, 6, 7]$$$.First operation:$$$i = 3$$$, $$$j = 5$$$, $$$k = 7$$$$$$x = 3$$$, $$$y = 5$$$, $$$z = 7$$$.$$$a = [1, 2, \operatorname{lcm}(y,z), 4, \operatorname{lcm}(x,z), 6, \operatorname{lcm}(x,y)]$$$ = $$$[1, 2, \color{red}{35}, 4, \color{red}{21}, 6, \color{red}{15}]$$$.Second operation:$$$i = 5$$$, $$$j = 6$$$, $$$k = 7$$$$$$x = 21$$$, $$$y = 6$$$, $$$z = 15$$$.$$$a = [1, 2, 35, 4, \operatorname{lcm}(y,z), \operatorname{lcm}(x,z), \operatorname{lcm}(x,y)]$$$ = $$$[1, 2, 35, 4, \color{red}{30}, \color{red}{105}, \color{red}{42}]$$$.Third operation:$$$i = 2$$$, $$$j = 3$$$, $$$k = 4$$$$$$x = 2$$$, $$$y = 35$$$, $$$z = 4$$$.$$$a = [1, \operatorname{lcm}(y,z), \operatorname{lcm}(x,z), \operatorname{lcm}(x,y), 30, 105, 42]$$$ = $$$[1, \color{red}{140}, \color{red}{4}, \color{red}{70}, 30, 105, 42]$$$.Subsequences whose GCD equal to $$$i$$$ is as follows:$$$\gcd(a_1, a_2) = \gcd(1, 140) = 1$$$$$$\gcd(a_3, a_4) = \gcd(4, 70) = 2$$$$$$\gcd(a_5, a_6, a_7) = \gcd(30, 105, 42) = 3$$$$$$\gcd(a_2, a_3) = \gcd(140, 4) = 4$$$$$$\gcd(a_2, a_4, a_5, a_6) = \gcd(140, 70, 30, 105) = 5$$$$$$\gcd(a_5, a_7) = \gcd(30, 42) = 6$$$$$$\gcd(a_2, a_4, a_6, a_7) = \gcd(140, 70, 105, 42) = 7$$$
The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^2$$$) — the number of test cases. The description of the test cases follows.The first and only line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 3 \cdot 10^{4}$$$) — the length of the array.It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^{4}$$$.
The first line should contain an integer $$$k$$$ ($$$0 \leq k \leq \lfloor \frac{n}{6} \rfloor + 5$$$) — where $$$k$$$ is the number of operations.The next $$$k$$$ lines should contain the description of each operation i.e. $$$3$$$ integers $$$i$$$, $$$j$$$ and $$$k$$$, where $$$1 \leq i, j, k \leq n$$$ and all must be distinct.
for _ in range(int(input())):
    (n, o) = (int(input()), [])
    if n % 4 in (0, 3):
        o.append((1, n, n - 1))
    (a, b) = (n - (3, 0, 1, 2)[n % 4], n - (2, 3, 0, 5)[n % 4])
    for i in range(a, n // 2, -4):
        o.append((i, i - 1, i - 2))
    if b // 2 >= o[-1][-1]:
        b -= 4
    for j in range(b, n // 3 * 2, -12):
        o.append((j, j - 4, j - 8))
    print(len(o))
    for x in o:
        print(*x)
import sys
ans1 = [8]
ans2 = [[[2, 6, 8], [3, 5, 7]]]
for _ in range(int(input())):
    n = int(input())
    if n in ans1:
        ans = ans2[ans1.index(n)]
        print(len(ans))
        for x in ans:
            print(' '.join(map(str, x)))
        continue
    ans = []
    pos = 0
    ost = []
    for i in range(3, n - 1, 4):
        if i > n // 2 - 2:
            ans.append([i, i + 1, i + 2])
            pos = i + 2
    for i in range(pos + 1, n + 1):
        if (i % 2 != 0 or i % 4 == 0) and i > n // 2:
            ost.append(i)
    per = n
    if (n - 1) % 4 == 2:
        per = n - 1
    elif (n - 2) % 4 == 2:
        per = n - 2
    elif (n - 3) % 4 == 2:
        per = n - 3
    for i in range(per, n // 2, -12):
        if i > n // 2:
            if i > 8:
                ans.append([i, i - 4, i - 8])
            else:
                ost.append(i)
    if len(ost) == 1:
        ans.append([1, 2, ost[0]])
    elif len(ost) == 2:
        ans.append([1, ost[1], ost[0]])
    elif len(ost) == 3:
        ans.append([ost[0], ost[1], ost[2]])
    elif len(ost) == 4:
        ans.append([1, ost[0], ost[1]])
        ans.append([2, ost[2], ost[3]])
    print(len(ans))
    for x in ans:
        print(' '.join(map(str, x)))
for _ in range(int(input())):
    n = int(input())
    (l, o, t) = (n // 2, [], [])
    if n % 4 in (0, 3):
        o.append((1, n, n - 1))
        n -= 2
    while n > l:
        if n % 2:
            o.append((n, n - 1, n - 2))
            n -= 3
        else:
            t.append(n)
            n -= 1
    if t and t[0] == 6:
        t = [6, 4, 2] if n == 5 else []
    while len(t) % 3:
        t.append(t[-1] - 4)
    for i in range(0, len(t), 3):
        o.append(t[i:i + 3])
    print(len(o))
    for x in o:
        print(*x)