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 ≤ i ≤ j ≤ n) — 

  * (p_i  OR  p_{i+1}  OR  …  OR  p_{j-1}  OR  p_{j}) ≥ j-i+1, where OR denotes the [bitwise OR operation.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)



In other words, a permutation p is good if for every subarray of p, the 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 ≤ t ≤ 100). Description of the test cases follows.

The first and only line of every test case contains a single integer n (1 ≤ n ≤ 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 OR 1 = 3 ≥ 2 (i = 1,j = 2) 
  * 3 OR 1 OR 2 = 3 ≥ 3 (i = 1,j = 3) 
  * 1 OR 2 = 3 ≥ 2 (i = 2,j = 3) 
  * 1 ≥ 1 (i = 2,j = 2) 



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

import math
import sys
import itertools
import random
import bisect
import heapq
import collections

def lcm(x, y):
	return (x*y)//math.gcd(x,y)
def hcf(x, y):
	return math.gcd(x, y)
def primes(n):
	prime=[True for _ in range(0, n+1)]
	res=[]
	p=2
	while p*p<=n:
		if prime[p]==True:
			for i in range(p*p,n+1,p):
				prime[p]=False
		p+=1
	for i in range(0, n+1):
		if prime[i]==True:
			res.append(i)
	return res
def primeFactors(n):
	res=[]
	while n%2==0:
		res.append(2)
		n//=2
	for i in range(3, math.sqrt(n)+1, 2):
		while n%i==0:
			res.append(i)
			n//=i
	if n>2:
		res.append(n)
	return res


for _ in range(0, int(input())):
	n=int(input())                                 # Single Digit Input
	# n, k = map(int, input().strip().split())    # Many Digit Input
	# l=list(map(int, input().strip().split()))      # List Input
	# s=list(input())                                # Character Array Input


	print(" ".join(str(x) for x in range(1, n+1)))