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;
  using vi = vector<int>;
  using vl = vector<ll>;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  void solve() {
    int n;
    cin >> n;
    vi a(n), freq(2);
    for (int& x : a) cin >> x, freq[x]++;
    if (freq[0] == 0) {
      cout << "NO\n";
      return;
    }
    vi ans;
    auto op = [&](int i) {
      ans.push_back(i);
      a[i] = a[i + 1] = a[i + 2] = a[i] ^ a[i + 1] ^ a[i + 2];
    };
    for (int z = 0; z < 2; z++) {
      int left = 0;
      for (int i = 0; i + 2 < n; i++) {
        if (a[i] + a[i + 1] + a[i + 2] == 2 && left % 2 == 0) {
          int idx = i;
          left += 2;
          while (left > 0) {
            op(idx);
            left -= 2;
            idx -= 2;
          }
          i++;
          continue;
        }
        if (i > 0 && a[i - 1] && a[i - 1] + a[i] + a[i + 1] == 1) op(i - 1);
        left += a[i];
      }
    }
    bool bad = ans.size() > n;
    for (int i : a)
      if (i) bad = 1;
    if (bad) {
      cout << "NO\n";
      return;
    }
    cout << "YES\n";
    cout << ans.size() << "\n";
    for (int i : ans) cout << i + 1 << " ";
    cout << "\n";
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  