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 pb push_back
//#define mp make_pair
#define fastread ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define openfile ifstream cin; ofstream cout; cin.open("input.txt"); cout.open("output.txt");
#define f(i, x, y) for(ll i = x; i < y; i++)
#define all(X) X.begin(), X.end()
#define int long long
#define ll long long
#define key pair<ll, ll>
#define keyy pair<pair<ll, ll>, ll>
#define keyyy pair<pair<ll, ll>, pair<ll, ll>>
// #define ordered_set<T> tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>
#define keyd pair<double, double>
#define ff first
#define ss second
#define double long double
const ll mod = 1e9 + 7;
// const ll mod = 998244353;
// const ll mod = 360 * 12 * (double)1e10;
const ll inf = 1e18 + 5;
const ll infi = 2e9;
using namespace std;

vector<int> fun(int n, int add = 0)
{
    vector<int> v;
    for(int i = 1; i <= n-2; i += 2)
        v.pb(i + add);
    for(int i = n-4; i > 0; i -= 2)
        v.pb(i + add);
    return v;
}

main()
{
    fastread;
    int T; cin>>T;
    while(T--)
    {
        int n; cin>>n; bool a[n+1]; f(i, 1, n+1) cin>>a[i];
        bool lft[n+2], rht[n+2]; lft[0] = rht[n+1] = 0;
        f(i, 1, n+1) lft[i] = (lft[i-1] + a[i]) % 2;
        // f(i, 1, n+1) cout<<lft[i]<<" "; cout<<"\n";
        if(lft[n] & 1) { cout<<"NO\n"; continue; }
        vector<int> ans;
        bool fl = true;
        if(n & 1)
            ans = fun(n);
        else
        {
            for(int i = 3; i < n-2; i += 2)
            {
                vector<int> v1, v2;
                if(lft[i]) continue;
                v1 = fun(i); v2 = fun(n-i, i);
                for(int j : v1) ans.pb(j);
                for(int j : v2) ans.pb(j);
                break;
            }
            if(a[1] == 0) ans = fun(n-1, 1);
            if(a[n] == 0) ans = fun(n-1);
        }
        f(i, 1, n+1)
            if(a[i] && (ans.size() == 0))
                fl = false;
        cout<<((fl)? "YES\n" : "NO\n");
        for(int i : ans) cout<<i<<" "; cout<<"\n";
    }
}