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
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__

def gift():
    for _ in range(t):
        n = int(input())
        
        ans = list(range(1,n+1 ))
        yield n
        yield " ".join([str(x) for x in ans])
if __name__ == '__main__':
    t= int(input())
    ans = gift()
    print(*ans,sep='\n')
            

#1+2
#"{} {} {}".format(maxele,minele,minele)
# 1 2 3 4
# 10 9 8 7

# 1 2 3 4

# 9 2 3 4
# 8 1 3 4
# 7 1 2 4
# 6 1 2 3

# 1 2 3
# 1 3 4
# 3 3 6
# 6 6 6