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 + 2;
  int N;
  int A[MAXN];
  int S[MAXN];
  void print_answer(int l, int r) {
    if (r - l < 3) {
      return;
    }
    for (int i = l; i <= r - 3; i += 2) {
      cout << i + 1 << ' ';
    }
    for (int i = r - 3; i >= l; i -= 2) {
      cout << i + 1 << ' ';
    }
  }
  void run_once() {
    cin >> N;
    int last_even_index_with_even_sum = -1;
    for (int i = 0; i < N; ++i) {
      cin >> A[i];
      S[i] = (i > 0) ? S[i - 1] : 0;
      S[i] += A[i];
      S[i] %= 2;
      if (i % 2 == 0 && S[i] == 0) {
        last_even_index_with_even_sum = i;
      }
    }
    if (N % 2 == 1) {
      if (S[N - 1] == 1) {
        cout << "NO\n";
      } else {
        cout << "YES\n" << N - 1 << "\n";
        print_answer(0, N);
        cout << '\n';
      }
    } else if (last_even_index_with_even_sum == -1 || S[N - 1] == 1) {
      cout << "NO\n";
    } else {
      cout << "YES\n" << N - 2 << "\n";
      print_answer(0, last_even_index_with_even_sum + 1);
      print_answer(last_even_index_with_even_sum + 1, N);
      cout << '\n';
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      run_once();
    }
    return 0;
  }
  