human
Alexandra has an even-length array a, consisting of 0s and 1s. The elements of the array are enumerated from 1 to n. She wants to remove at most n/2 elements (where n — length of array) in the way that alternating sum of the array will be equal 0 (i.e. a_1 - a_2 + a_3 - a_4 + ... = 0). In other words, Alexandra wants sum of all elements at the odd positions and sum of all elements at the even positions to become equal. The elements that you remove don't have to be consecutive.

For example, if she has a = [1, 0, 1, 0, 0, 0] and she removes 2nd and 4th elements, a will become equal [1, 1, 0, 0] and its alternating sum is 1 - 1 + 0 - 0 = 0.

Help her!

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^3, n is even) — length of the array.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — elements of the array.

It is guaranteed that the sum of n over all test cases does not exceed 10^3.

Output

For each test case, firstly, print k (n/2 ≤ k ≤ n) — number of elements that will remain after removing in the order they appear in a. Then, print this k numbers. Note that you should print the numbers themselves, not their indices.

We can show that an answer always exists. If there are several answers, you can output any of them. 

Example

Input

4
2
1 0
2
0 0
4
0 1 1 1
4
1 1 0 0

Output

1
0
1
0
2
1 1
4
1 1 0 0

Note

In the first and second cases, alternating sum of the array, obviously, equals 0.

In the third case, alternating sum of the array equals 1 - 1 = 0.

In the fourth case, alternating sum already equals 1 - 1 + 0 - 0 = 0, so we don't have to remove anything.
Solve the task in PYTHON3.
 Write me a Python code to print k (n/2 ≤ k ≤ n) — number of elements that will remain after removing in the order they appear in a?
assistant
#Here is Python code to print k (n/2 ≤ k ≤ n) — number of elements that will remain after removing in the order they appear in a.

#!/usr/bin/env python
# https://github.com/cheran-senthil/PyRival/blob/master/templates/template_py3.py
import os
import sys,math
from io import BytesIO, IOBase

def main():
    t=int(input())
    for case in range(t):
        n=int(input())
        l=list(map(int,input().split()))  
        ct0 = l.count(0); ct1 = l.count(1)
        if ct0>(n//2):
            while ct0%2!=0: ct0-=1
            print(ct0)
            for i in range(ct0): print(0,end=" ")
            print()
        elif ct1>(n//2):
            while ct1%2!=0: ct1-=1
            print(ct1)
            for i in range(ct1): print(1,end=" ")
            print()
        else:
            ct0=n//2
            print(ct0)
            for i in range(ct0): print(0,end=" ")
            print()
            # check = ((n//2)%2==0)
            # print(n//2+(0 if check else 1))
            # for i in range(n):
            #     if ct1==0 and check: break
            #     if l[i]==0:
            #         if check: continue
            #         check = True
            #         print(0,end=" ")
            #     else:
            #         ct1-=1
            #         print(1,end=" ")
            # print()

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr =...