Problem: There are n cats in a line, labeled from 1 to n, with the i-th cat at position i. They are bored of gyrating in the same spot all day, so they want to reorder themselves such that no cat is in the same place as before. They are also lazy, so they want to minimize the total distance they move. Help them decide what cat should be at each location after the reordering.

For example, if there are 3 cats, this is a valid reordering: [3, 1, 2]. No cat is in its original position. The total distance the cats move is 1 + 1 + 2 = 4 as cat 1 moves one place to the right, cat 2 moves one place to the right, and cat 3 moves two places to the left.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

The first and only line of each test case contains one integer n (2 ≤ n ≤ 100) — the number of cats.

It can be proven that under the constraints of the problem, an answer always exist. 

Output

Output t answers, one for each test case. Each answer consists of n integers — a permutation with the minimum total distance. If there are multiple answers, print any.

Example

Input


2
2
3


Output


2 1 
3 1 2 

Note

For the first test case, there is only one possible permutation that satisfies the conditions: [2, 1].

The second test case was described in the statement. Another possible answer is [2, 3, 1].

Solution (in PYTHON3):

Solution:
# import sys
# sys.stdin=open('input.txt','r')
# sys.stdout=open('output.txt','w')
import copy
# snippets gcd lcm snippet .............................................................

def gcd(x, y):
    while (y):
        x, y = y, x % y
    return x

def lcm(x,y):
    return x*y//gcd(x,y)


# for alphagoai

def gcd(a, b):
    while (b):
        a, b = b, a % b
    return a

def lcm(a,b):
    return a*b//gcd(a,b)
# to check number is prime or not in n**.5//6 complexity

def is_prime(n):
    if (n <= 1):
        return False
    if (n <= 3):
        return True
    if (n % 2 == 0 or n % 3 == 0):
        return False
    i = 5
    while (i * i <= n):
        if (n % i == 0 or n % (i + 2) == 0):
            return False
        i = i + 6

    return True


# kmp search snippet ...........................................................


def LSPtable(pattern):
    n=len(pattern)
    l=[0]*n
    j=0
    for i in range(1,n):
        while j>0 and pattern[i]!=pattern[j]:
            j=l[j-1]
        if pattern[i]==pattern[j]:
            l[i]=j+1
            j+=1
        else:
            l[i]=0
    return l

def KMPsearch(pattern,string):
    lsp=LSPtable(pattern)
    j=0
    for i in range(len(string)):
        while j>0 and string[i]!=pattern[j]:
            j=lsp[j-1]

        if string[i]==pattern[j]:
            j+=1
            if j== len(pattern):
                return i-j+1
    return -1

# Binary Exponentiation
# takes logarithmic time complex for calculatation of a power of some watch erritcho video if forgot
# below is iterative implementation of binary exponentiation as iterative's are faster than recurssive
def power(a,b):
    res=1
    while b>0:
        if b%2:res*=a
        a*=a
        b=b//2
    return res

# mod inverse function......................
#     if m is prime ....... fermats's little theorem

def modInverse(a, m):
    g = gcd(a, m)

    if (g != 1):
        print("Inverse doesn't exist")

    else:
        print("Modular multiplicative inverse is ",
              power(a, m - 2, m))

def power(x, y, m):
    if (y == 0):
        return 1

    p = power(x, y // 2, m) % m
    p = (p * p) % m

    if (y % 2 == 0):
        return p
    else:
        return ((x * p) % m)

def gcd(a, b):
    if (a==0):
        return b

    return gcd(b % a, a)

# fenwick tree snippet .........................................................

def getsum(BITTree,i):
    s=0
    while i>0:
        s+=BITTree[i]
        i-=i&(-i)
    return s

def updatebit(BITTree,n,i,v):
    i=i+1
    while i<=n:
        BITTree[i]+=v
        i+=i&(-i)

def constructor(arr, n):
    BITTree =[0]*(n+1)
    for i in range(n):
        updatebit(BITTree,n,i,arr[i])
    return BITTree

def arrIn():
    return list(map(int,input().split()))
def mapIn():
    return map(int,input().split())

# import sys
# sys.stdin=open('input.txt','r')
# sys.stdout=open('output.txt','w')


for i in range(int(input())):
    n=int(input())
    if n%2==0:
        x=1
        y=2
        for i in range(n):
            if i%2:
                print(x,end=" ")
                x+=2
            else:
                print(y,end=" ")
                y+=2
        print()
    else:
        x=1
        y=2
        for i in range(n-3):
            if i%2:
                print(x,end=" ")
                x+=2
            else:
                print(y,end=" ")
                y+=2
        print(n,n-2,n-1)