A permutation of length n is an array of size n consisting of n distinct integers in the range [1, n]. For example, (3, 2, 4, 1) is a permutation of length 4, but (3, 3, 1, 4) and (2, 3, 4, 5) are not, as (3, 3, 1, 4) contains duplicate elements, and (2, 3, 4, 5) contains elements not in range [1,4].
A permutation p of length n is good if and only if for any 1 ≤ i ≤ n, pi ≠ i.
Please find the lexicographically smallest good permutation p.
Definition for "lexicographically smaller":
For two permutations p and q, we say that p is lexicographically smaller than q if and only if there exists a index 1 ≤ l ≤ n such that:
- For any 1 ≤ i < l, pi = qi. Note that if l = 1, this constraint means nothing.
- and, pl < ql.

For example, (2, 3, 1, 4) < (2, 3, 4, 1) < (3, 4, 1, 2). The lexicographically smallest permutation is, of course, (1, 2, ..., n), though this one is not good.

-----Input-----
First line of the input contains an integer T denoting number of test cases.
For each test case, the only line contains an integer n.

-----Output-----
For each test case, output the lexicographically smallest good permutation of length n. It's guaranteed that this permutation exists.

-----Constraints-----
- 1 ≤ T ≤ 10
- 2 ≤ n ≤ 105

-----Subtasks-----
- Subtask #1 (17 points): 2 ≤ n ≤ 9
- Subtask #2 (83 points): Original Constraints

-----Example-----
Input:
4
2
3
5
6

Output:
2 1
2 3 1
2 1 4 5 3
2 1 4 3 6 5

-----Explanation-----
Example case 1. The only good permutation of length 2 is (2, 1).
Example case 2. Consider all permutations of length 3, they are(in lexicographically order):
- p = (1, 2, 3), it's not good since p[1] = 1, p[2] = 2 and p[3] = 3;
- p = (1, 3, 2), it's not good since p[1] = 1;
- p = (2, 1, 3), it's not good since p[3] = 3;
- p = (2, 3, 1), it's good since p[1] ≠ 1, p[2] ≠ 2 and p[3] ≠ 3;
- p = (3, 1, 2), it's good since p[1] ≠ 1, p[2] ≠ 2 and p[3] ≠ 3;
- p = (3, 2, 1), it's not good since p[2] = 2.
Thus the minimum good one is (2, 3, 1).
Example case 3. Consider two good permutations for third test case: p=(2, 1, 4, 5, 3) and q=(2, 4, 1, 5, 3), then p < q. You can check lexicographically condition as follows. Find the first index where the entries of two permutations are different, and compare those entries. For example, in this case, the first position where the entries differ is index 2. You can see that p[2] < q[2], as 1 < 4, so p is lexicographically smaller than q.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = "# cook your dish here\ntry:\n    for _ in range(int(input())):\n        n=int(input())\n        l=[]\n        if (n%2==0):\n            for i in range(1,n+1):\n                if(i%2==0):\n                    l.append(str(i-1))\n                else:\n                    l.append(str(i+1))\n        else:\n            for j in range(1,n-1):\n                if(j%2==0):\n                    l.append(str(j-1))\n                else:\n                    l.append(str(j+1))\n            l.append(str(n))\n            l.append(str(n-2))\n        print(' '.join(l))\nexcept:\n    pass\n\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13447