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].
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))