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.
import sys
# from collections import deque
# from collections import Counter
# from math import sqrt
# from math import log
# from math import ceil
# from bisect import bisect_left, bisect_right

# alpha=['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
# mod=10**9+7
# mod=998244353

# def BinarySearch(a,x): 
# 	i=bisect_left(a,x) 
# 	if(i!=len(a) and a[i]==x): 
# 		return i 
# 	else: 
# 		return -1

# def sieve(n): 
# 	prime=[True for i in range(n+1)]
# 	p=2
# 	while(p*p<=n): 
# 		if (prime[p]==True): 
# 			for i in range(p*p,n+1,p): 
# 				prime[i]=False
# 		p+=1
# 	prime[0]=False
# 	prime[1]=False
# 	s=set()
# 	for i in range(len(prime)):
# 		if(prime[i]):
# 		s.add(i)
# 	return s

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

fast_reader=sys.stdin.readline
fast_writer=sys.stdout.write

def input():
	return fast_reader().strip()

def print(*argv):
	fast_writer(' '.join((str(i)) for i in argv))
	fast_writer('\n')

#____________________________________________________________________________________________________________________________________

for _ in range(int(input())):
	n=int(input())
	print(*[i+1 for i in range(n)])