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>
#include <unistd.h>
#include <time.h>
#include <sys/time.h>

using namespace std;

///////////////////////////////////////////////////////////////////////////////

#define DEBUG 0

#define pb push_back
#define V vector
#define S static

#define rep(i,n) for(ll i=0LL; i<n; ++i)
#define srep(i,s,n) for(ll i=s; i<n; ++i)
#define rrep(i,n) for(ll i=n-1LL; i>=0LL; --i)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define CIN(x) do { \
        assert(!cin.eof()); \
        cin >> x; \
        assert(!cin.fail()); \
} while(0);
#if DEBUG
#define debug_print(...) _debug_print(__VA_ARGS__)
#define debug_printf(...) printf(__VA_ARGS__)
#define debug_print_time _debug_print_time
#else // DEBUG
#define debug_print(...)
#define debug_printf(...)
#define debug_print_time
#endif // DEBUG

typedef long long ll;
typedef unsigned long long ull;
typedef __int128_t ll128;
typedef tuple<ll, ll> t2;
typedef tuple<ll, ll, ll> t3;
typedef tuple<ll, ll, ll, ll> t4;
typedef tuple<ll, ll, ll, ll, ll> t5;
template <typename T>
using priority_queue_incr = priority_queue<T, V<T>, greater<T> >;

///////////////////////////////////////////////////////////////////////////////

template <typename TT, typename T>
void incr_m(map<TT, T> &m, TT k)
{
    if (m.find(k) == m.end()) m[k] = 0;
    m[k]++;
}

void llin(ll &a)
{
    CIN(a);
}

void llinl1(V<ll> &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a;
        CIN(a);
        v.push_back(a);
    }
}

void llinl2(V<t2> &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a, b;
        CIN(a >> b);
        v.push_back(t2(a, b));
    }
}

void llinl3(V<t3> &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a, b, c;
        CIN(a >> b >> c);
        v.push_back(t3(a, b, c));
    }
}

void llinl4(V<t4> &v, ll count)
{
    for (ll i = 0LL; i < count; ++i) {
        ll a, b, c, d;
        CIN(a >> b >> c >> d);
        v.push_back(t4(a, b, c, d));
    }
}

void llina(V<ll> &v, ll count)
{
    llinl1(v, count);
}

template <typename T>
void sort(V<T> &v)
{
    sort(v.begin(), v.end());
}

template <typename T>
void sort_reverse(V<T> &v)
{
    sort(v.begin(), v.end(), greater<T>());
}

t2 _ext_gcd(ll a, ll b, ll g)
{
    if (!b) return t2(1, 0);
    ll q = a / b;
    ll r = a % b;
    t2 sol = _ext_gcd(b, r, g);
    ll sx = get<0>(sol);
    ll sy = get<1>(sol);
    ll x = sy;
    ll y = sx - q * sy;
    return t2(x, y);
}

t2 ext_gcd(ll a, ll b)
{
    return _ext_gcd(a, b, gcd(a, b));
}

// x and mod must be coprime
ll mod_inv(ll x, ll mod)
{
    t2 result = ext_gcd(x, mod);
    ll ret = get<0>(result);
    while (ret < 0) ret += mod;
    ret %= mod;
    return ret;
}

ll msec()
{
    struct timeval tv;
    gettimeofday(&tv, NULL);
    ll ret = 0;
    ret += tv.tv_sec * 1000LL;
    ret += tv.tv_usec / 1000LL;
    return ret;
}


template <typename T>
void _debug_print(T x)
{
    cout << x << " ";
}

void _debug_print(const t2 &x)
{
    ll x1 = get<0>(x);
    ll x2 = get<1>(x);
    cout << "(" << x1 << ", " << x2 << ") ";
}

void _debug_print(const t3 &x)
{
    ll x1 = get<0>(x);
    ll x2 = get<1>(x);
    ll x3 = get<2>(x);
    cout << "(" << x1 << ", " << x2 << ", " << x3 << ") ";
}

void _debug_print(const t4 &x)
{
    ll x1 = get<0>(x);
    ll x2 = get<1>(x);
    ll x3 = get<2>(x);
    ll x4 = get<3>(x);
    cout << "(" << x1 << ", " << x2 << ", " << x3 << ", " << x4 << ") ";
}

template <typename T>
void _debug_print(T xarray[], ll n)
{
    rep (i, n) _debug_print(xarray[i]);
    cout << endl;
}

template <typename T>
void _debug_print(const V<T> &xlist)
{
    for (auto x : xlist) _debug_print(x);
    cout << endl;
}

template <typename T>
void _debug_print(const set<T> &xset)
{
    for (auto x : xset) _debug_print(x);
    cout << endl;
}

template <typename TT, typename T>
void _debug_print(const map<TT, T> &xlist)
{
    for (auto x : xlist) {
        TT k = x.first;
        T v = x.second;
        cout << "K=";
        _debug_print(k);
        cout << " V=";
        _debug_print(v);
        cout << endl;
    }
}

///////////////////////////////////////////////////////////////////////////////

void _main();
int main()
{
    cout << setprecision(12);
#if !DEBUG
    ios::sync_with_stdio(false);
    cin.tie(0);
#endif

    _main();
    return 0;
}

// * cause of WA
// ** map<ll, ll> maps; ll val = maps[non-existent-key];
// ** memory initialization: ll costs[10]; memset(costs, 0, sizeof(bool) * 10);
// ** skip enqueue when graph BFS/DFS
// * cause of TLE
// ** rep (i, too_many_cnt) debug_printf(....);
// ** for (auto s : slist) { ...; } <-- each s is big string
// ** rep (i, many_cnt) cout << ans[i] << endl;
//       --> [good] string ansstr; rep (i, many_cnt) ansstr += ans[i] + "\n"; cout << ansstr;

bool ng(V<ll> alist, ll n)
{
    ll cnt = 0;
    for (const auto & a :alist) if (a) cnt++;
    if (cnt & 1LL) {
        cout << "NO" << endl;
        return true;
    }
    if (cnt == n) {
        cout << "NO" << endl;
        return true;
    }

    ll cnts[2] = {};
    rep (i, n) {
        if (alist[i]) cnts[0]++;
        else break;
    }
    rrep (i, n) {
        if (alist[i]) cnts[1]++;
        else break;
    }
    if ((cnts[0] & 1LL) && (cnts[1] & 1LL) && cnts[0] + cnts[1] == cnt) {
        cout << "NO" << endl;
        return true;
    }

    return false;
}

void op(V<ll> &ans, V<ll> &alist, ll n, ll i)
{
    debug_printf("op %lld\n", i);
    if (i == 0) return;
    if (i == n-1) return;
    if (alist[i-1] == alist[i] && alist[i] == alist[i+1]) return;
    ll cnt = 0;
    cnt += alist[i-1];
    cnt += alist[i];
    cnt += alist[i+1];
    if (cnt & 1LL) alist[i-1] = alist[i] = alist[i+1] = 1;
    else alist[i-1] = alist[i] = alist[i+1] = 0;
    debug_printf("result %lld\n", alist[i]);
    ans.pb(i);
}

void solve()
{
    ll n; llin(n);
    V<ll> alist; llina(alist, n);

    debug_printf("----------------------------- solve\n");
    debug_print(alist);

    if (ng(alist, n)) return;

    V<ll> ans;
    ll idx0 = -1;

    rep (i, n) {
        if (alist[i]) continue;
        if (!(i & 1LL)) {
            idx0 = i;
            break;
        }
        op(ans, alist, n, i);
        if (alist[i]) continue;
        idx0 = i-1;
        break;
    }
    if (idx0 < 0) {
        cout << "NO" << endl;
        return;
    }
    assert(!(idx0 & 1LL));
    debug_printf("idx0=%lld\n", idx0);
    debug_printf("alist 1\n");
    debug_print(alist);

    cout << "YES" << endl;

    for (; idx0; idx0 -= 2) op(ans, alist, n, idx0-1);
    debug_printf("alist 2\n");
    debug_print(alist);

    assert(!alist[0]);
    rep (i, n) {
        if (!alist[i]) continue;
        op(ans, alist, n, i+1);
        if (!alist[i]) continue;
        op(ans, alist, n, i);
    }

    cout << ans.size() << endl;
    for (const auto &x : ans) cout << x << " ";
    if (ans.size()) cout << endl;
}

void _main()
{
    ll t; llin(t);
    rep (i, t) solve();
}

///////////////////////////////////////////////////////////////////////////////
