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 w, n, xo, pos;
  vector<int> ans;
  int main() {
    ios_base::sync_with_stdio(false);
    cin >> w;
    while (w--) {
      cin >> n;
      for (int i = 1; i <= n; i++) {
        int x;
        cin >> x;
        xo ^= x;
        if (xo == 0 && i % 2 == 1) {
          pos = i;
        }
      }
      if (xo == 0 && pos) {
        for (int i = 1; i < pos; i += 2) {
          ans.push_back(i);
        }
        for (int i = pos - 4; i >= 1; i -= 2) {
          ans.push_back(i);
        }
        if (n % 2 == 0) {
          pos++;
        }
        for (int i = pos; i + 2 <= n; i += 2) {
          ans.push_back(i);
        }
        for (int i = n - 4; i >= pos; i -= 2) {
          ans.push_back(i);
        }
        cout << "YES\n";
        cout << ans.size() << "\n";
        for (auto x : ans) {
          cout << x << " ";
        }
        cout << "\n";
      } else {
        cout << "NO\n";
      }
      xo = 0, pos = 0, ans.clear();
    }
    return 0;
  }
  