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 solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    int s = 0;
    for (auto &i : a) cin >> i, s += i;
    if (s & 1) {
      cout << "NO\n";
      return;
    }
    int pos = -1;
    for (int i = 0; i < n; ++i) {
      if (a[i] == 0 && s % 2 == 0) pos = i;
      s += a[i];
    }
    vector<int> ans;
    if (pos == -1) {
      for (int i = 0; i < n; ++i)
        if (a[i] == 1) {
          int j = i + 1;
          while (j < n && a[j] == 0) ++j;
          if (j < n && (j - i) % 2 == 0) {
            int k = i;
            while (k + 2 < j) {
              ans.push_back(k);
              a[k + 1] = a[k + 2] = 1;
              k += 2;
            }
            ans.push_back(k);
            a[k] = a[k + 2] = 0;
            pos = k + 1;
            break;
          }
          i = j - 1;
        }
    }
    if (pos == -1) {
      cout << "NO\n";
      return;
    }
    for (int i = pos; i > 1; --i) {
      if (a[i - 1] == 0) continue;
      if (a[i - 2] == 1) {
        ans.push_back(i - 2);
        a[i - 1] = a[i - 2] = 0;
        continue;
      }
      int j = i - 2;
      while (j >= 0 && a[j] == 0) --j;
      if (j < 0) break;
      int k = i - 1;
      while (k - 2 > j) {
        k -= 2;
        ans.push_back(k);
        a[k] = a[k + 1] = 1;
      }
      if (k != j + 1) {
        ans.push_back(j);
        a[j] = a[j + 2] = 0;
      }
      ++i;
    }
    for (int i = pos; i + 2 < n; ++i) {
      if (a[i + 1] == 0) continue;
      if (a[i + 2] == 1) {
        ans.push_back(i);
        a[i + 1] = a[i + 2] = 0;
        continue;
      }
      int j = i + 2;
      while (j < n && a[j] == 0) ++j;
      if (j >= n) break;
      int k = i + 1;
      while (k + 2 < j) {
        ans.push_back(k);
        a[k + 1] = a[k + 2] = 1;
        k += 2;
      }
      if (k != j - 1) {
        ans.push_back(j - 2);
        a[j - 2] = a[j] = 0;
      }
      --i;
    }
    assert(*max_element(a.begin(), a.end()) == 0);
    cout << "YES\n";
    cout << ans.size() << "\n";
    for (auto &i : ans) cout << i + 1 << " \n"[&i == &ans.back()];
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    int tc = 1;
    cin >> tc;
    while (tc--) solve();
    return 0;
  }
  