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;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      int db = 0;
      vector<int> v(n), ki;
      int ind = -1, ossz = 0;
      for (int i = 0; i < n; i++) {
        cin >> v[i];
        ossz += v[i];
        if ((ossz & 1) == 0 && v[i] == 0) ind = i;
        if (ind == -1) {
          if (i > 2 && v[i - 1] == 0 && v[i - 2] == 0) {
            v[i - 2] = v[i - 1] = 1;
            ossz += 2;
            ki.push_back(i - 2);
          } else if (i > 1 && v[i - 1] == 0 && v[i - 2] == 1 && v[i] == 1) {
            v[i - 2] = v[i] = 0;
            ossz -= 2;
            ki.push_back(i - 1);
            ind = i - 1;
          }
        }
      }
      if ((ossz & 1) || ind == -1) {
        cout << "NO\n";
        continue;
      }
      for (int i = ind; i > 1; i--) {
        if (v[i - 1] == 1 && v[i - 2] == 1) {
          ki.push_back(i - 1);
          v[i - 1] = v[i - 2] = 0;
        } else if (v[i - 1] == 1 && v[i - 2] == 0) {
          ki.push_back(i - 2);
          if (v[i - 3] == 0) {
            ki.push_back(i - 1);
            v[i - 1] = 0;
            v[i - 3] = 1;
          } else {
            v[i - 1] = v[i - 3] = 0;
          }
        }
      }
      for (int i = ind; i < n - 2; i++) {
        if (v[i + 1] == 1 && v[i + 2] == 1) {
          ki.push_back(i + 1);
          v[i + 1] = v[i + 2] = 0;
        } else if (v[i + 1] == 1 && v[i + 2] == 0) {
          ki.push_back(i + 2);
          if (v[i + 3] == 0) {
            ki.push_back(i + 1);
            v[i + 1] = 0;
            v[i + 3] = 1;
          } else {
            v[i + 1] = v[i + 3] = 0;
          }
        }
      }
      cout << "YES\n";
      cout << ki.size() << '\n';
      for (int i = 0; i < (int)ki.size(); i++)
        cout << ki[i] << " \n"[i == (int)ki.size() - 1];
    }
    return 0;
  }
  