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() {
    long long n, su = 0;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; ++i) {
      cin >> a[i];
      su += a[i];
    }
    if (su % 2) {
      cout << "NO\n";
      return;
    }
    long long s = 0, ind = -1;
    for (long long i = 0; i < n; ++i) {
      if (!a[i] && s % 2 == 0) {
        ind = i;
        break;
      }
      s += a[i];
    }
    vector<long long> ans;
    if (ind == -1) {
      for (long long i = n - 2; i > 0; i -= 2) {
        if (a[i]) {
          i++;
          continue;
        }
        ans.push_back(i);
        a[i] = 1;
        if (a[i - 1]) {
          ind = i - 1;
          a[i - 1] = a[i] = a[i + 1] = 0;
          break;
        }
        a[i - 1] = 1;
      }
      if (ind == -1) {
        cout << "NO\n";
        return;
      }
    }
    for (long long i = ind - 1; i >= 0; --i) {
      if (a[i]) {
        if (a[i - 1]) {
          ans.push_back(i);
          a[i - 1] = a[i] = 0;
        } else {
          if (a[i - 2]) {
            ans.push_back(i - 1);
            a[i - 2] = a[i] = 0;
          } else {
            ans.push_back(i - 1);
            ans.push_back(i);
            a[i - 2] = 1;
            a[i] = 0;
          }
        }
      }
    }
    for (long long i = ind + 1; i < n; ++i) {
      if (a[i]) {
        if (a[i + 1]) {
          ans.push_back(i);
          a[i + 1] = a[i] = 0;
        } else {
          if (a[i + 2]) {
            ans.push_back(i + 1);
            a[i + 2] = a[i] = 0;
          } else {
            ans.push_back(i + 1);
            ans.push_back(i);
            a[i + 2] = 1;
            a[i] = 0;
          }
        }
      }
    }
    cout << "YES\n" << ans.size() << '\n';
    for (long long i = 0; i < ans.size(); ++i) {
      cout << ans[i] << " ";
    }
    cout << '\n';
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  