human
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.
Using python3 can you solve the prior task?
assistant
#Here is the Python code you requested.

import sys
import math
import copy
import heapq
from fractions import *
from decimal import *
from functools import cmp_to_key
from bisect import bisect_left, bisect_right
from collections import defaultdict, deque, Counter

# sys.setrecursionlimit(1000000)

# input aliases
input = sys.stdin.readline
getS = lambda: input().strip()
getN = lambda: int(input())
getList = lambda: list(map(int, input().split()))
getZList = lambda: [int(x) - 1 for x in input().split()]

INF = float("inf")
MOD = 10 ** 9 + 7
divide = lambda x: pow(x, MOD - 2, MOD)

def nck(n, k, kaijyo):
    return (npk(n, k, kaijyo) * divide(kaijyo[k])) % MOD

def npk(n, k, kaijyo):
    if k == 0 or k == n:
        return n % MOD
    return (kaijyo[n] * divide(kaijyo[n - k])) % MOD

def fact_and_inv(SIZE):
    inv = [0] * SIZE  # inv[j] = j^{-1} mod MOD
    fac = [0] * SIZE  # fac[j] = j! mod MOD
    finv = [0] * SIZE  # finv[j] = (j!)^{-1} mod MOD
    inv[1] = 1
    fac[0] = fac[1] = 1
    finv[0] = finv[1] = 1
    for i in range(2, SIZE):
        inv[i] = MOD - (MOD // i) * inv[MOD % i] % MOD
        fac[i] = fac[i - 1] * i % MOD
        finv[i] = finv[i - 1] * inv[i] % MOD

    return fac, finv

def renritsu(A, Y):
    # example 2x + y = 3, x + 3y = 4
    # A = [[2,1], [1,3]])
    # Y = [[3],[4]] または [3,4]
    A = np.matrix(A)
    Y = np.matrix(Y)
    Y = np.reshape(Y, (-1, 1))
    X = np.linalg.solve(A, Y)

    # [1.0, 1.0]
    return X.flatten().tolist()[0]

class TwoDimGrid:
    # 2次元座標 -> 1次元
    def __init__(self, h, w, wall="#"):
        self.h = h
        self.w = w
        self.size = (h + 2) * (w + 2)
        self.wall = wall
        self.get_grid()
        # self.init_cost()

    def get_grid(self):
        grid = [self.wall * (self.w + 2)]
        for i in range(self.h):
            grid.append(self.wall + getS() + self.wall)

        grid.append(self.wall * (self.w + 2))
        self.grid = grid

    def init_cost(self):
        self.cost = [INF] * self.size

    def pos(self, x, y):
        # 壁も含めて0-indexed 元々の座標だけ考えると1-indexed
...