A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array) and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).

For a positive integer $n$, we call a permutation $p$ of length $n$ good if the following condition holds for every pair $i$ and $j$ ($1 \le i \le j \le n$) —   $(p_i \text{ OR } p_{i+1} \text{ OR } \ldots \text{ OR } p_{j-1} \text{ OR } p_{j}) \ge j-i+1$, where $\text{OR}$ denotes the bitwise OR operation. 

In other words, a permutation $p$ is good if for every subarray of $p$, the $\text{OR}$ of all elements in it is not less than the number of elements in that subarray. 

Given a positive integer $n$, output any good permutation of length $n$. We can show that for the given constraints such a permutation always exists.


-----Input-----

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). Description of the test cases follows.

The first and only line of every test case contains a single integer $n$ ($1 \le n \le 100$).


-----Output-----

For every test, output any good permutation of length $n$ on a separate line. 


-----Example-----
Input
3
1
3
7

Output
1
3 1 2
4 3 5 2 7 1 6



-----Note-----

For $n = 3$, $[3,1,2]$ is a good permutation. Some of the subarrays are listed below.   $3\text{ OR }1 = 3 \geq 2$ $(i = 1,j = 2)$  $3\text{ OR }1\text{ OR }2 = 3 \geq 3$ $(i = 1,j = 3)$  $1\text{ OR }2 = 3 \geq 2$ $(i = 2,j = 3)$  $1 \geq 1$ $(i = 2,j = 2)$ 

Similarly, you can verify that $[4,3,5,2,7,1,6]$ is also good.

Difficulty: interview

Starter code:

Solutions:
for _ in range(int(input())):
    n = int(input())
    print(*list(range(1, n+1)))


# for _ in range(1):
for _ in range(int(input())):
    # a, b = map(int, input().split())
    n = int(input())
    # arr = list(map(int, input().split()))
    # s = input()
    for i in range(n):
        print(i + 1, end=' ')
    print()


import sys
import math
def II():
	return int(sys.stdin.readline())

def LI():
	return list(map(int, sys.stdin.readline().split()))

def MI():
	return map(int, sys.stdin.readline().split())

def SI():
	return sys.stdin.readline().strip()
t = II()
for q in range(t):
	n = II()
	l = list(range(1,n+1))
	l= l[:][::-1]
	print(*l)

for _ in range(int(input())):
    n = int(input())
    print(*list(range(1,n+1)))

from math import *
from collections import *
from random import *
from decimal import Decimal
from heapq import *
from bisect import *
import sys
input=sys.stdin.readline
sys.setrecursionlimit(10**5)
def lis():
    return list(map(int,input().split()))
def ma():
    return map(int,input().split())
def inp():
    return int(input())
def st1():
    return input().rstrip('\n')
t=inp()
while(t):
    t-=1
    n=inp()
    for i in range(1,n+1):
        print(i,end=' ')
    print()


a=int(input())
for i in range(a):
    s=int(input())
    z=[i+1 for i in range(s)]
    print(*z)
    


'''input
3
1
3
7
'''
import math	
def solve():
	n = int(input())
	l = [i for i in range(1,n+1)]
	l = l[::-1]
	for i in l:
		print(i,end=" ")
	print()
t = 1
t = int(input())
while t > 0:
	t-=1
	solve()

import math
t = int(input())
for test in range(t):
    n = int(input())
    for i in range(1,n+1):
        print(i,end=" ")
    print()    


from sys import stdin
input = stdin.readline

for _ in range(int(input())):
    n = int(input())

    for x in range(1, n + 1):
        print(x, end = ' ')

    print()

from collections import *
from sys import stdin,stderr
def rl():
    return [int(w) for w in stdin.readline().split()]

t, = rl()
for _ in range(t):
    n, = rl()
    print(*list(range(1, n+1)))