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].
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define lli long long int
#define INF 1000000000000000000
#define llu unsigned long long int
#define pb push_back
#define ff first
#define ss second
long long mod=1000000007;
/*
bool comparator(const pair<vector<int>,int> a,const pair<vector<int>,int> b)
{
return a.ff.size()<b.ff.size();
}
*/
vector<int> getans(int n,int add)
{
  int i;
  vector <int> ans;
  for(i=1;i<n;i+=2)
  ans.pb(i+add);
  for(i=n-4;i>=1;i--)
  ans.pb(i+add);
  return ans;
}
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t,n,i,f,pos;
cin>>t;
while(t--)
{
  cin>>n;
  vector <int> a(n+1),pref(n+1);
  for(i=1;i<=n;i++)
  cin>>a[i];
  pref[1]=a[1];
  for(i=2;i<=n;i++)
  pref[i]=pref[i-1]^a[i];
  f=0;
  if(pref[n]!=0)
  f=1;
  if(n%2==0)
  {
    f=1;pos=-1;
    for(i=1;i<n;i+=2)
    if(pref[i]==0)
    {
      pos=i;
      f=0;
      break;
    }
  }
  if(f)
  cout<<"NO\n";
  else
  {
    cout<<"YES\n";
    vector <int> ans;
    if(n%2)
    ans=getans(n,0);
    else
    {
      ans=getans(n-i,i);
      if(i!=1)
      {
        vector <int> tmp=getans(i,0);
        n=(int)tmp.size();
        for(i=0;i<n;i++)
        ans.pb(tmp[i]);
      }
    }
    n=(int)ans.size();
    cout<<n<<"\n";
    for(i=0;i<n;i++)
    cout<<ans[i]<<" ";
    cout<<"\n";
  }
}
}