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;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  clock_t startTime;
  double getCurrentTime() {
    return (double)(clock() - startTime) / CLOCKS_PER_SEC;
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    int totalXor = 0;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      totalXor ^= a[i];
    }
    if (totalXor == 1) {
      cout << "NO\n";
      return;
    }
    auto solve = [](int l, int r) {
      assert((r - l + 1) % 2 == 1);
      vector<int> res;
      for (int i = l; i <= r - 2; i += 2) {
        res.push_back(i);
      }
      for (int i = r - 4; i >= l; i -= 2) {
        res.push_back(i);
      }
      return res;
    };
    if (n % 2 == 1) {
      auto res = solve(0, n - 1);
      cout << "YES\n";
      cout << res.size() << "\n";
      for (auto x : res) {
        cout << x + 1 << " ";
      }
      cout << "\n";
    } else {
      bool ok = false;
      int prefXor = 0;
      for (int i = 0; i < n; ++i) {
        prefXor ^= a[i];
        if (prefXor == 0 && i % 2 == 0) {
          auto res1 = solve(0, i);
          auto res2 = solve(i + 1, n - 1);
          cout << "YES\n";
          cout << res1.size() + res2.size() << "\n";
          for (auto x : res1) {
            cout << x + 1 << " ";
          }
          for (auto x : res2) {
            cout << x + 1 << " ";
          }
          cout << "\n";
          ok = true;
          break;
        }
      }
      if (!ok) cout << "NO\n";
    }
  }
  int main() {
    startTime = clock();
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int tests;
    cin >> tests;
    for (int t = 0; t < tests; ++t) {
      solve();
    }
    return 0;
  }
  