Problem: There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end. 

To achieve this, you will:

  * Choose m such that 1 ≤ m ≤ 1000

  * Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.




Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies. 

  * It can be proved that for the given constraints such a sequence always exists.

  * You don't have to minimize m.

  * If there are several valid sequences, you can output any.

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 each test case contains one integer n (2 ≤ n≤ 100). 

Output

For each testcase, print two lines with your answer. 

In the first line print m (1≤ m ≤ 1000) — the number of operations you want to take. 

In the second line print m positive integers a_1, a_2, ..., a_m (1 ≤ a_i ≤ n), where a_j is the number of bag you chose on the j-th operation.

Example

Input


2
2
3


Output


1
2
5
3 3 3 1 2

Note

In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.

In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag — and the result is [12, 12, 12].

Solution (in PYTHON3):

Solution:
import sys
from math import *
from collections import Counter as cc , defaultdict as df , OrderedDict as od 
inf = float('inf')
def counter(l): return dict(cc(l))
def lcm(a,b): 
    return int((a*b) / gcd(a,b))
def yes(): print('YES')
def no(): print('NO') 
def so(x):      return {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
def rev(l):     return list(reversed(l))
def ini():      return int(sys.stdin.readline())
def inp():      return map(int, sys.stdin.readline().strip().split())
def li():       return list(map(int, sys.stdin.readline().strip().split()))
def i1():    return sys.stdin.readline().strip()
for i in range(ini()):
    n=ini()
    print(n)
    print(*[i for i in range(1,n+1)])