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>
  using namespace std;
  using ll = long long;
  using pii = pair<int, int>;
  template <class T>
  int size(T &&a) {
    return (int)(a.size());
  }
  ostream &operator<<(ostream &os, string str) {
    for (char c : str) os << c;
    return os;
  }
  template <class A, class B>
  ostream &operator<<(ostream &os, const pair<A, B> &p) {
    return os << '(' << p.first << "," << p.second << ')';
  }
  template <class T>
  auto operator<<(ostream &os, T &&x) -> decltype(x.begin(), os) {
    os << '{';
    for (auto it = x.begin(); it != x.end(); ++it)
      os << *it << (it == prev(x.end()) ? "" : " ");
    return os << '}';
  }
  template <class T>
  ostream &operator<<(ostream &os, vector<vector<T>> vec) {
    for (auto x : vec) os << "\n  " << x;
    return os;
  }
  void dump() {}
  template <class T, class... Args>
  void dump(T &&x, Args... args) {
    cerr << x << "; ";
    dump(args...);
  }
  vector<int> ans;
  void print_sequence(int st, int en) {
    if (en - st + 1 < 3) return;
    for (int i = st; i <= en - 2; i += 2) ans.emplace_back(i);
    for (int i = en - 4; i >= st; i -= 2) ans.emplace_back(i);
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n), pref_xor(n);
    for (int i = 0; i < (n); ++i) {
      cin >> a[i];
      if (i == 0)
        pref_xor[i] = a[i];
      else
        pref_xor[i] = pref_xor[i - 1] ^ a[i];
    }
    if (pref_xor[n - 1] == 1) {
      cout << "NO\n";
      return;
    }
    ans.clear();
    if (n % 2 == 1) {
      cout << "YES\n";
      print_sequence(0, n - 1);
      cout << size(ans) << '\n';
      for (int x : ans) cout << x + 1 << ' ';
      cout << '\n';
      return;
    }
    for (int i = 0; i < n; i += 2) {
      if (pref_xor[i] == 0) {
        cout << "YES\n";
        print_sequence(0, i);
        print_sequence(i + 1, n - 1);
        cout << size(ans) << '\n';
        for (int x : ans) cout << x + 1 << ' ';
        cout << '\n';
        return;
      }
    }
    cout << "NO\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) solve();
  }
  