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?

Can you provide a the Python code you requested?
assistant
#Here is the Python code you requested.

from sys import stdin
from collections import deque
# https://codeforces.com/contest/1354/status/D
mod = 10**9 + 7
import sys
import random
sys.setrecursionlimit(10**6)
from queue import PriorityQueue
from collections import Counter as cc
# def rl():
#     return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
from itertools import permutations
from math import factorial as f

# def ncr(x, y):
#     return f(x) // (f(y) * f(x - y))
def ncr(n, r, p):
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den,
                      p - 2, p)) % p

import sys

# input = sys.stdin.readline
# LCA
# def bfs(na):
#
#     queue = [na]
#     boo[na] = True
#     level[na] = 0
#
#     while queue!=[]:
#
#         z = queue.pop(0)
#
#         for i in hash[z]:
#
#             if not boo[i]:
#
#                 queue.append(i)
#                 level[i] = level[z] + 1
#                 boo[i] = True
#                 dp[i][0] = z
#
#
#
# def prec(n):
#
#     for i in range(1,20):
#
#         for j in range(1,n+1):
#             if dp[j][i-1]!=-1:
#                 dp[j][i] = dp[dp[j][i-1]][i-1]
#
#
# def lca(u,v):
#     if level[v] < level[u]:
#         u,v = v,u
#
#     diff = level[v] - level[u]
#
#
#     for i in range(20):
#         if ((diff>>i)&1):
#             v = dp[v][i]
#
#
#     if u == v:
#         return u
#
#
#     for i in range(19,-1,-1):
#         # print(i)
#         if dp[u][i] != dp[v][i]:
#
#             u = dp[u][i]
#             v = dp[v][i]
#
#
#     return dp[u][0]
#
# dp = []
#
#
# n =...