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.
"""Template for Python Competitive Programmers prepared by pa.n.ik, kabeer seth and Mayank Chaudhary """

# to use the print and division function of Python3
from __future__ import division, print_function

"""value of mod"""
MOD = 998244353
mod = 10**9 + 7

"""use resource"""
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])

"""for factorial"""

def prepare_factorial():
    fact = [1]
    for i in range(1, 100005):
        fact.append((fact[-1] * i) % mod)
    ifact = [0] * 100005
    ifact[100004] = pow(fact[100004], mod - 2, mod)
    for i in range(100004, 0, -1):
        ifact[i - 1] = (i * ifact[i]) % mod

    return fact, ifact

"""uncomment next 4 lines while doing recursion based question"""
# import threading
# threading.stack_size(1<<27)
import sys
# sys.setrecursionlimit(10000)


"""uncomment modules according to your need"""

from bisect import bisect_left, bisect_right, insort
from math import floor, ceil, sqrt, degrees, atan, pi, log, sin, radians
from heapq import heappop, heapify, heappush
from collections import Counter, defaultdict, deque


def modinv(n, p):
    return pow(n, p - 2, p)

def ncr(n, r, fact, ifact):    # for using this uncomment the lines calculating fact and ifact
    t = (fact[n] * (ifact[r] * ifact[n-r]) % mod) % mod
    return t

def get_ints(): return map(str, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()

"""*****************************************************************************************"""

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

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

def get_xor(n):
    return [n, 1, n+1, 0][n % 4]

def fast_expo(a, b):
    res = 1
    while b:
        if b&1:
            res = (res * a)
            res %= MOD
            b -= 1
        else:
            a = (a* a)
            a %= MOD
            b>>=1
    res %= MOD
    return res

def get_n(P):    # this function returns the maximum n for which Summation(n) <= Sum

    ans = (-1 + sqrt(1 + 8*P))//2
    return ans

""" ********************************************************************************************* """

def main():

    T = int(input())
    while T:
        n = int(input())
        Arr = [i for i in range(1, n+1)]
        print(*Arr)
        T -= 1


""" -------- Python 2 and 3 footer by Pajenegod and c1729 ---------"""

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192

class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):

    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

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

""" main function"""

if __name__ == '__main__':
    main()
    # threading.Thread(target=main).start()