description: 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].
solution: #include <bits/stdc++.h>
  #pragma GCC optimize("O3")
  #pragma GCC target("sse4")
  using namespace std;
  template <class T>
  using pq = priority_queue<T>;
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  void __print(int x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char* x) { cerr << '\"' << x << '\"'; }
  void __print(const string& x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V>& x) {
    cerr << '{';
    __print(x.first);
    cerr << ", ";
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T& x) {
    int first = 0;
    cerr << '{';
    for (auto& i : x) cerr << (first++ ? ", " : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  const int MOD = 1000000007;
  const char nl = '\n';
  const int MX = 100001;
  void solve() {
    int N;
    cin >> N;
    int A[N];
    for (int i = 0; i < (N); i++) cin >> A[i];
    int sum = 0;
    for (int i = 0; i < (N); i++) sum += A[i];
    if (sum % 2) {
      cout << "NO" << nl;
      return;
    }
    int cnt = 0;
    vector<int> ops;
    for (int i = 0; i < (N); i++) {
      cnt += A[i];
      if (A[i] == 0) {
        if (cnt % 2) {
          ops.push_back(i - 1);
          int X = 0;
          for (int j = 0; j < (3); j++) X ^= A[i - 1 + j];
          for (int j = 0; j < (3); j++) A[i - 1 + j] = X;
        }
        if (A[i] == 0) {
          cnt = 0;
        } else
          cnt++;
      }
    }
    sum = 0;
    for (int i = 0; i < (N); i++) sum += A[i];
    if (sum == N) {
      cout << "NO" << nl;
      return;
    }
    int fz = 0;
    for (int i = (N)-1; i >= 0; i--)
      if (A[i] == 0) fz = i;
    while (fz > 0) {
      ops.push_back(fz - 2);
      A[fz - 2] = 0;
      A[fz - 1] = 0;
      A[fz] = 0;
      fz -= 2;
    }
    for (int i = 0; i < (N - 1); i++) {
      if (A[i] == 0 && A[i + 1] == 1) {
        ops.push_back(i);
        A[i] = 0;
        A[i + 1] = 0;
        A[i + 2] = 0;
      }
    }
    cout << "YES" << nl;
    cout << (int)(ops).size() << nl;
    for (auto& a : ops) {
      cout << a + 1 << " ";
    }
    cout << nl;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  