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;
  const int maxn = 2e5 + 10;
  int N, A[maxn];
  vector<int> ans;
  bool rev;
  int f(int i) { return (A[i] + A[i + 1] + A[i + 2]) & 1; }
  void op(int i) {
    if (i + 2 > N) return;
    if (!rev)
      ans.emplace_back(i);
    else
      ans.emplace_back(N - i + 1 - 2);
    A[i] = A[i + 1] = A[i + 2] = f(i);
  }
  void go(int i) {
    if (i > N) return;
    if (!A[i]) {
      go(i + 1);
      return;
    }
    int j = i;
    while (j <= N && A[j] == 1) ++j;
    if (j > N) return;
    if ((j - i) % 2 == 0) {
      for (int k = j - 2; k >= 1 && A[k] && A[k + 1]; k -= 2) op(k);
      go(j);
      return;
    }
    if (j + 1 <= N && A[j + 1] == 1) {
      op(j - 1);
      for (int k = j - 3; k >= 1 && A[k] && A[k + 1]; k -= 2) op(k);
      go(j + 2);
      return;
    }
    op(j - 1);
    go(j + 1);
  }
  void solve() {
    cin >> N;
    A[N + 1] = A[N + 2] = 1;
    for (int i = 1; i <= N; ++i) cin >> A[i];
    if (count(A + 1, A + 1 + N, 1) & 1) {
      cout << "NO" << '\n';
      return;
    }
    ans.clear();
    rev = false;
    go(1);
    reverse(A + 1, A + 1 + N);
    rev = true;
    go(1);
    reverse(A + 1, A + 1 + N);
    rev = false;
    go(1);
    for (int i = 1; i <= N; ++i)
      if (A[i]) {
        cout << "NO" << '\n';
        return;
      }
    for (auto& x : ans)
      if (x + 2 > N) {
        cout << "NO" << '\n';
        return;
      }
    cout << "YES" << '\n';
    cout << ans.size() << '\n';
    copy((ans).begin(), (ans).end(), ostream_iterator<int>(cout, " ")),
        cout << '\n';
  }
  int main(void) {
    ios_base::sync_with_stdio(0), cin.tie(0);
    int TC;
    cin >> TC;
    while (TC--) {
      solve();
    }
    return 0;
  }
  