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;
  void solve() {
    int n, sum = 0;
    ;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      sum += a[i];
    }
    if (sum % 2 != 0) {
      cout << "NO\n";
    } else {
      vector<int> ones, ops;
      for (int i = 0; i < n; ++i) {
        if (a[i] == 1) {
          ones.push_back(i);
        } else {
          if (((int)ones.size()) % 2 == 0) {
            while (!ones.empty()) {
              ones.pop_back();
              ops.push_back(ones.back());
              ones.pop_back();
            }
          } else if (i < n - 1) {
            if (a[i + 1] == 1) {
              ops.push_back(ones.back());
              ones.pop_back();
              while (!ones.empty()) {
                ones.pop_back();
                ops.push_back(ones.back());
                ones.pop_back();
              }
              ++i;
            } else {
              ops.push_back(ones.back());
              ones.push_back(i);
              ones.push_back(i + 1);
              ++i;
            }
          }
        }
      }
      if ((int)ones.size() == n || ((int)ones.size()) % 2 != 0) {
        cout << "NO\n";
      } else if ((int)ones.size() > 0) {
        for (int i = 0; i < (int)ones.size(); i += 2) {
          ops.push_back(ones[i] - 1);
        }
        ones.clear();
      }
      if ((int)ones.size() == 0) {
        cout << "YES\n";
        cout << (int)ops.size() << '\n';
        for (int op : ops) {
          cout << op + 1 << ' ';
        }
        cout << '\n';
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  