description: You are given a sequence a of length n consisting of 0s and 1s.
  
  You can perform the following operation on this sequence: 
  
    * Pick an index i from 1 to n-2 (inclusive). 
    * Change all of a_{i}, a_{i+1}, a_{i+2} to a_{i} ⊕ a_{i+1} ⊕ a_{i+2} simultaneously, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
  
  Find a sequence of at most n operations that changes all elements of a to 0s or report that it's impossible.
  
  We can prove that if there exists a sequence of operations of any length that changes all elements of a to 0s, then there is also such a sequence of length not greater than n.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4).
  
  The first line of each test case contains a single integer n (3 ≤ n ≤ 2⋅10^5) — the length of a.
  
  The second line of each test case contains n integers a_1, a_2, …, a_n (a_i = 0 or a_i = 1) — elements of a.
  
  It is guaranteed that the sum of n over all test cases does not exceed 2⋅10^5.
  
  Output
  
  For each test case, do the following: 
  
    * if there is no way of making all the elements of a equal to 0 after performing the above operation some number of times, print "NO". 
    * otherwise, in the first line print "YES", in the second line print k (0 ≤ k ≤ n) — the number of operations that you want to perform on a, and in the third line print a sequence b_1, b_2, ..., b_k (1 ≤ b_i ≤ n - 2) — the indices on which the operation should be applied. 
  
  
  
  If there are multiple solutions, you may print any.
  
  Example
  
  Input
  
  
  3
  3
  0 0 0
  5
  1 1 1 1 0
  4
  1 0 0 1
  
  
  Output
  
  
  YES
  0
  YES
  2
  3 1
  NO
  
  Note
  
  In the first example, the sequence contains only 0s so we don't need to change anything.
  
  In the second example, we can transform [1, 1, 1, 1, 0] to [1, 1, 0, 0, 0] and then to [0, 0, 0, 0, 0] by performing the operation on the third element of a and then on the first element of a.
  
  In the third example, no matter whether we first perform the operation on the first or on the second element of a we will get [1, 1, 1, 1], which cannot be transformed to [0, 0, 0, 0].
solution: import sys,io,os
  try:Z=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
  except:Z=lambda:sys.stdin.readline().encode()
  Y=lambda:map(int,Z().split())
  add=lambda x:x+1
  
  def f(n,a):
      m=f=1;c=g=0;x=n;xc=0;z=[]
      for i in range(n):
          #print(m,c,f,g,x,xc,z)
          if a[i]==m:c+=1
          else:
              if m==1:
                  if c&1==0:
                      for j in range(c//2):z.append(i-2*j-2)
                      if f:f=g=0
                  else:g=1;x=i;xc=c
                  c=1
              else:
                  if g:
                      if c&1==0:
                          if f:
                              for j in range(x-1,i-2,2):z.append(j)
                              c=i+1
                          else:
                              for j in range(x-xc-1,x-3,2):z.append(j)
                              for j in range(x-2,i-3,2):z.append(j)
                              z.append(i-2)
                              for j in range(i-4,x-3,-2):z.append(j)
                              c=g=0
                      else:
                          if f:f=0
                          for j in range(x-1,i-3,2):z.append(j)
                          z.append(i-2)
                          for j in range(i-4,x-2,-2):z.append(j)
                          for j in range(x-3,x-xc-1,-2):z.append(j)
                          c=g=0
                  else:
                      if f:f=0
                      c=1
          m=a[i]
      if g or (m and (c&1 or c==n)):return 0,[]
      if m:
          for j in range(n-c-1,n-2,2):z.append(j)
      return 1,z
  
  O=[]
  for _ in range(int(Z())):
      b,z=f(int(Z()),[*Y()])
      if b:
          O.append("YES");O.append(str(len(z)))
          if z:O.append(' '.join(map(str,map(add,z))))
      else:O.append("NO")
  print('\n'.join(O))