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;
  using ll = long long;
  using ld = long double;
  template <class T>
  bool uin(T& a, const T& b) {
    return a > b ? a = b, 1 : 0;
  }
  template <class T>
  bool uax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  void solve() {
    int n;
    cin >> n;
    int cur = 0;
    vector<int> a(n);
    for (auto& x : a) {
      cin >> x;
      cur ^= x;
    }
    if (cur != 0) return void(cout << "NO\n");
    vector<int> ops;
    auto do_op = [&](int i) {
      assert(0 <= i && i < n - 2);
      int x = a[i] ^ a[i + 1] ^ a[i + 2];
      a[i] = a[i + 1] = a[i + 2] = x;
      ops.push_back(i);
    };
    auto solve = [&](int l, int r) {
      assert((r - l) % 2);
      for (int i = l; i < r - 2; i += 2) do_op(i);
      for (int i = r - 5; i >= l; i -= 2) do_op(i);
    };
    if (n & 1)
      solve(0, n);
    else {
      int j = 1;
      cur = a[0];
      for (; j + 1 < n; j += 2) {
        if (cur == 0) break;
        cur ^= a[j] ^ a[j + 1];
      }
      if (cur != 0) return void(cout << "NO\n");
      solve(0, j);
      solve(j, n);
    }
    for (auto x : a) assert(x == 0);
    cout << "YES\n";
    cout << int((ops).size()) << '\n';
    for (auto x : ops) cout << x + 1 << ' ';
    cout << '\n';
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) solve();
  }
  