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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const int MaxN = 500100;
  int N;
  vector<int> a;
  bool pref[MaxN];
  int breakpoint;
  void check(vector<int> num, const vector<int> &ops) {
    for (auto mid : ops) {
      int r = num[mid - 1] ^ num[mid] ^ num[mid + 1];
      for (int z = mid - 1; z <= mid + 1; z++) num[z] = r;
    }
    for (int i = (1); i < (N + 1); i++) assert(num[i] == 0);
  }
  bool conditions() {
    if (pref[N] != 0) return false;
    for (int i = (1); i < (N + 1); i++) {
      if (i % 2 == 1 && pref[i] == 0) {
        breakpoint = i;
        return true;
      }
    }
    return false;
  }
  void solve() {
    if (!conditions()) {
      cout << "NO" << endl;
      return;
    };
    if (N % 2 == 1) breakpoint = N;
    assert(breakpoint % 2 == 1);
    vector<int> ops;
    for (int i = 2; i <= breakpoint - 1; i += 2) {
      ops.push_back(i);
    }
    for (int i = breakpoint - 3; i >= 2; i -= 2) {
      ops.push_back(i);
    }
    for (int i = breakpoint + 2; i <= N - 1; i += 2) {
      ops.push_back(i);
    }
    for (int i = N - 3; i >= breakpoint + 2; i -= 2) {
      ops.push_back(i);
    }
    cout << "YES" << endl;
    cout << ops.size() << endl;
    for (auto mid : ops) {
      cout << mid - 1 << " ";
    }
    cout << endl;
    check(a, ops);
  }
  void load() {
    breakpoint = -1;
    cin >> N;
    a.clear();
    a.resize(N + 1);
    a[0] = -100;
    for (int i = (0); i < (N); i++) cin >> a[i + 1];
    pref[0] = 0;
    for (int i = (0); i < (N); i++) pref[i + 1] = pref[i] ^ a[i + 1];
  }
  int main() {
    ios_base::sync_with_stdio(false);
    int T;
    cin >> T;
    for (int tt = (0); tt < (T); tt++) {
      load();
      solve();
    }
    return 0;
  }
  