You are given an integer $D$. Find an integer sequence $A_1, A_2, \ldots, A_N$ such that the following conditions are satisfied:
- $1 \le N \le 10^5$
- $1 \le A_i \le 10^5$ for each valid $i$
- $\sum_{i=1}^N \sum_{j=i}^N \left( \mathrm{min}(A_i, A_{i+1}, \ldots, A_j) - \mathrm{GCD}(A_i, A_{i+1}, \ldots, A_j) \right) = D$
It can be proved that a solution always exists under the given constraints.
Note: $\mathrm{GCD}(B_1, B_2, \ldots, B_M)$ is the greatest integer which divides all the integers $B_1, B_2, \ldots, B_M$.
-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first and only line of each test case contains a single integer $D$.
-----Output-----
For each test case, print two lines. The first of these lines should contain a single integer $N$. The second line should contain $N$ space-separated integers $A_1, A_2, \ldots, A_N$.
If there are multiple solutions, you may find any one of them.
-----Constraints-----
- $1 \le T \le 10$
- $0 \le D \le 10^9$
-----Example Input-----
4
2
5
200
13
-----Example Output-----
3
3 3 2 
5
2 8 5 1 10 
7
12 10 15 11 19 13 15
4
5 4 4 10
# cook your dish here
def solve(d):
 op=[]
 p=(10**5)-2
 if d>0:
  while d>0:
   p=min(d,p)
   op.append(p+1)
   op.append(p+2)
   op.append(1)
   d= d-p
 else:
   op.append(d+1)
   op.append(d+2)
 print(len(op))
 print(*op)
def __starting_point():
 for test in range(int (input())):
  d=int(input())
  solve(d)
__starting_point()
# cook your dish here
def solve(d):
 op=[]
 p=(10**5)-2
 if d>0:
  while d>0:
   p=min(d,p)
   op.append(p+1)
   op.append(p+2)
   op.append(1)
   d= d-p
 else:
   op.append(d+1)
   op.append(d+2)
 print(len(op))
 print(*op)
def __starting_point():
 for test in range(int (input())):
  d=int(input())
  solve(d)
__starting_point()