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>

#define ll long long
#define int ll
#define f first
#define s second
#define pb push_back
#define sz(x) (int)x.size()
#define all(s) s.begin(), s.end()

using namespace std;


const int N = (int)2e5+100;
const int M = (int)3e5 + 100;
const int inf = (int)2e9;
int mod = (int)998244353;
const long long INF = (int)1e18;

int n, a[N];
bool pref[N];
vector<int>ans;

void Do(int i){
  int s=(a[i]^a[i+1]^a[i+2]);
  a[i]=a[i+1]=a[i+2]=s;
  ans.pb(i);
}

void solve(){
  ans.clear();
  cin>>n;
  for(int i=1;i<=n;i++) cin>>a[i],pref[i]=(pref[i-1]^a[i]);
  if(pref[n]){
    cout<<"NO\n";
    return;
  }
  if(n&1){
    for(int i=1;i<n;i+=2) Do(i);
    for(int i=n-2;i>=1;i-=2) Do(i);
  }else{
    for(int i=1;i<=n;i++){
      if(i%2==1&&!pref[i]&&!(pref[n]^pref[i])){
        for(int j=1;j<i;j+=2) Do(j);
        for(int j=i-2;j>=1;j-=2) Do(j);
        for(int j=i+1;j<n;j+=2) Do(j);
        for(int j=n-1;j>i;j-=2) Do(j);
        break;
      }
    }
  }
  for(int i=1;i<=n;i++){
    if(a[i]){
      cout<<"NO\n";
      return;
    }
  }
  cout<<"YES\n"<<sz(ans)<<"\n";
  for(auto v:ans)cout<<v<<"\n";   
}


main () {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin>>t;
  while(t--)solve();
}