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].
//雪花飄飄北風嘯嘯
//天地一片蒼茫

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#define ll long long
#define ii pair<ll,ll>
#define iii pair<ii,ll>
#define fi first
#define se second
#define endl '\n'
#define debug(x) cout << #x << ": " << x << endl

#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define lb lower_bound
#define ub upper_bound

#define rep(x,start,end) for(auto x=(start)-((start)>(end));x!=(end)-((start)>(end));((start)<(end)?x++:x--))
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()

#define indexed_set tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>
//change less to less_equal for non distinct pbds, but erase will bug

mt19937 rng(chrono::system_clock::now().time_since_epoch().count());

const int MOD=998244353;

ll qexp(ll b,ll p,int m){
    ll res=1;
    while (p){
        if (p&1) res=(res*b)%m;
        b=(b*b)%m;
        p>>=1;
    }
    return res;
}

ll inv(ll i){
	return qexp(i,MOD-2,MOD);
}

ll fix(ll i){
	i%=MOD;
	if (i<0) i+=MOD;
	return i;
}

ll fac[1000005];
ll ifac[1000005];

ll nCk(int i,int j){
	if (i<j) return 0;
	return fac[i]*ifac[j]%MOD*ifac[i-j]%MOD;
}

int n;
int arr[200005];

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin.exceptions(ios::badbit | ios::failbit);
	
	fac[0]=1;
	rep(x,1,1000005) fac[x]=fac[x-1]*x%MOD;
	ifac[1000004]=inv(fac[1000004]);
	rep(x,1000005,1) ifac[x-1]=ifac[x]*x%MOD;
	
	int TC;
	cin>>TC;
	while (TC--){
		cin>>n;
		rep(x,0,n) cin>>arr[x];
		
		vector<int> ans;
		vector<ii> v;
		
		int p=-1;
		
		int cnt=0;
		rep(x,0,n) if (arr[x]){
			cnt^=1;
			if (p!=-1 && cnt==0) v.pub(ii(p,x));
			p=x;
		}
		
		//for (auto &it:v) cout<<it.fi<<" "<<it.se<<endl;
		
		
		for (auto &it:v){
			for (int y=it.fi;y<it.se-1-(it.se-it.fi+1)%2;y+=2){
				arr[y+1]=arr[y+2]=1;
				ans.pub(y);
			}
		}
		
		cnt=0;
		
		rep(x,0,n-2){
			if (arr[x]) cnt^=1;
			
			if (cnt && arr[x]+arr[x+2]==2 && arr[x+1]==0){
				ans.pub(x);
				arr[x]=arr[x+1]=arr[x+2]=0;
			}
		}
		
		//rep(x,0,n) cout<<arr[x]<<" "; cout<<endl;
		
		rep(x,0,n-2){
			if (arr[x]+arr[x+1]+arr[x+2]==2){
				ans.pub(x);
				arr[x]=arr[x+1]=arr[x+2]=0;
			}
		}
		
		rep(x,n-2,0){
			if (arr[x]+arr[x+1]+arr[x+2]==2){
				ans.pub(x);
				arr[x]=arr[x+1]=arr[x+2]=0;
			}
		}
		
		bool can=true;
		rep(x,0,n) if (arr[x]) can=false;
		
		if (can){
			cout<<"YES"<<endl;
			cout<<sz(ans)<<endl;
			for (auto &it:ans) cout<<it+1<<" "; cout<<endl;
		}
		else{
			cout<<"NO"<<endl;
		}
	}
}
