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 calc(vector<int> &arr, vector<int> &ids, int l, int r) {
    for (int i = r - 2; i >= l; i -= 2) {
      ids.push_back(i);
      int v = arr[i] ^ arr[i + 1] ^ arr[i + 2];
      arr[i] = arr[i + 1] = arr[i + 2] = v;
    }
    for (int i = l; i <= r - 2; i += 2) {
      ids.push_back(i);
      int v = arr[i] ^ arr[i + 1] ^ arr[i + 2];
      arr[i] = arr[i + 1] = arr[i + 2] = v;
    }
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> arr(n + 1), pre(n + 1), ids;
    for (int i = 1; i <= n; i++) {
      cin >> arr[i];
      pre[i] = pre[i - 1] ^ arr[i];
    }
    if (pre[n] != 0) {
      cout << "NO\n";
      return;
    }
    if (n & 1)
      calc(arr, ids, 1, n);
    else {
      bool f = false;
      for (int i = 1; !f && i <= n; i += 2) {
        if (pre[i] == 0) {
          f = true;
          calc(arr, ids, 1, i);
          calc(arr, ids, i + 1, n);
        }
      }
      if (!f) {
        cout << "NO\n";
        return;
      }
    }
    cout << "YES\n";
    if (ids.empty()) {
      cout << "0\n";
      return;
    }
    cout << (int)ids.size() << "\n";
    for (int id : ids) cout << id << " ";
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t-- > 0) solve();
    return 0;
  }
  