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")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  template <class T>
  inline void amin(T &x, const T &y) {
    if (y < x) x = y;
  }
  template <class T>
  inline void amax(T &x, const T &y) {
    if (x < y) x = y;
  }
  int N;
  int A[200011];
  void MAIN() {
    scanf("%d", &N);
    for (int i = 0, i_len = (N); i < i_len; ++i) scanf("%d", A + i);
    int cnt = count(A, A + N, 1);
    bool yes = true;
    if (cnt % 2) yes = false;
    if (yes) {
      if (N % 2 == 0 && A[0] == 1 && A[N - 1] == 1) {
        bool p = true;
        for (int i = 1; i + 1 < N; i += 2) {
          if (A[i] != A[i + 1]) p = false;
        }
        if (p) yes = false;
      }
    }
    if (yes) {
      vector<int> ans;
      for (int i = 0; i < N - 2;) {
        int m = A[i] + A[i + 1] + A[i + 2];
        if (m == 0) {
          i++;
        } else if (m == 1) {
          if (A[i] == 1) {
            ans.push_back(i + 1);
            A[i] = A[i + 1] = A[i + 2] = 1;
          } else {
            i++;
          }
        } else if (m == 2) {
          ans.push_back(i + 1);
          A[i] = A[i + 1] = A[i + 2] = 0;
          i = max(0, i - 2);
        } else if (m == 3) {
          if (i && A[i - 1] == 0)
            i--;
          else
            i += 2;
        } else {
          assert(false);
        }
      }
      assert(count(A, A + N, 1) == 0);
      puts("YES");
      printf("%d\n", (int)ans.size());
      if (ans.size()) {
        do {
          const auto end_rp = (ans.end());
          auto it_rp = (ans.begin());
          for (bool sp_rp = 0; it_rp != end_rp; ++it_rp) {
            if (sp_rp)
              putchar(' ');
            else
              sp_rp = true;
            printf("%d", *it_rp);
          }
          putchar('\n');
        } while (0);
      }
    } else {
      puts("NO");
    }
  }
  int main() {
    int TC = 1;
    scanf("%d", &TC);
    for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
    return 0;
  }
  