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 v[200005];
  vector<int> rz;
  vector<int> pzs;
  void upd(int pz) {
    int val = (v[pz] ^ v[pz + 1] ^ v[pz + 2]);
    v[pz] = v[pz + 1] = v[pz + 2] = val;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t, i, n, m, r, s, j, q;
    cin >> t;
    while (t--) {
      cin >> n;
      int cnt = 0;
      for (i = 1; i <= n; ++i) {
        cin >> v[i];
        cnt += v[i];
      }
      if (cnt & 1) {
        cout << "NO\n";
        continue;
      }
      pzs.clear();
      rz.clear();
      for (i = 1; i <= n; ++i) {
        if (v[i]) pzs.push_back(i);
      }
      v[0] = v[n + 1] = 1;
      for (i = 0; i < pzs.size(); i += 2) {
        int a = pzs[i];
        int b = pzs[i + 1];
        if ((b - a) % 2 == 0) {
          for (j = a; j < b; j += 2) {
            upd(j);
            rz.push_back(j);
          }
          for (j = b - 4; j >= a; j -= 2) {
            upd(j);
            rz.push_back(j);
          }
        } else {
          if (v[a - 1] == 0) {
            for (j = a; j < b - 1; j += 2) {
              upd(j);
              rz.push_back(j);
            }
            for (j = a - 1; j <= b - 2; j += 2) {
              upd(j);
              rz.push_back(j);
            }
          }
        }
      }
      int ok = 1;
      for (i = pzs.size() - 1; i >= 0; i -= 2) {
        int b = pzs[i];
        int a = pzs[i - 1];
        if (v[a] == 0) continue;
        assert((b - a) % 2 == 1);
        if (v[b + 1] != 0) ok = 0;
        for (j = b - 2; j > a; j -= 2) {
          upd(j);
          rz.push_back(j);
        }
        for (j = b - 1; j >= a; j -= 2) {
          upd(j);
          rz.push_back(j);
        }
      }
      if (!ok) {
        cout << "NO\n";
        continue;
      }
      cout << "YES\n";
      cout << rz.size();
      cout << '\n';
      for (i = 0; i < rz.size(); ++i) cout << rz[i] << ' ';
      cout << '\n';
    }
    return 0;
  }
  