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 vi = vector<long long>;
  template <typename T>
  std::istream& operator>>(std::istream& input, std::pair<T, T>& data) {
    input >> data.first >> data.second;
    return input;
  }
  template <typename T>
  std::istream& operator>>(std::istream& input, std::vector<T>& data) {
    for (T& first : data) input >> first;
    return input;
  }
  template <typename T>
  std::ostream& operator<<(std::ostream& output, const pair<T, T>& data) {
    output << "(" << data.first << "," << data.second << ")";
    return output;
  }
  template <typename T>
  std::ostream& operator<<(std::ostream& output, const std::vector<T>& data) {
    for (const T& first : data) output << first << " ";
    return output;
  }
  std::ostream& operator<<(std::ostream& output, const __int128& first) {
    __int128 n = first;
    if (n == 0) {
      output << "0";
      return output;
    }
    if (n < 0) {
      n = -n;
      output << "-";
    }
    string s;
    while (n) {
      s += '0' + (n % 10);
      n /= 10;
    }
    reverse(s.begin(), s.end());
    cout << s;
    return output;
  }
  long long div_up(long long a, long long b) {
    return a / b + ((a ^ b) > 0 && a % b);
  }
  long long div_down(long long a, long long b) {
    return a / b - ((a ^ b) < 0 && a % b);
  }
  long long math_mod(long long a, long long b) { return a - b * div_down(a, b); }
  template <class T>
  using V = vector<T>;
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  long long gcd(long long a, long long b) {
    while (b) {
      tie(a, b) = make_pair(b, a % b);
    }
    return a;
  }
  long long Bit(long long mask, long long bit) { return (mask >> bit) & 1; }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long t;
    cin >> t;
    for (long long _ = (0); _ < (t); ++_) {
      long long n;
      cin >> n;
      vi a(n);
      cin >> a;
      long long sum = 0;
      for (auto& e : a) {
        sum += e;
      }
      if (sum & 1) {
        cout << "NO" << '\n';
        continue;
      }
      long long pos = -1, l = 0;
      for (long long i = (0); i < (n); ++i) {
        if ((!a[i]) && l % 2 == 0) {
          pos = i;
          break;
        }
        l += a[i];
      }
      auto make = [&](long long mid) {
        a[mid - 1] = a[mid] = a[mid + 1] = a[mid - 1] ^ a[mid] ^ a[mid + 1];
      };
      vi ans;
      if (pos == -1) {
        long long cur = 0;
        for (long long i = (0); i < (n); ++i) {
          if (a[i]) {
            if (cur & 1) {
              for (long long j = 1; j < cur; j += 2) {
                make(i - j);
                ans.push_back(i - j);
              }
              make(i - cur);
              ans.push_back(i - cur);
              pos = i - cur;
              break;
            }
            cur = 0;
          } else {
            cur++;
          }
        }
      };
      if (pos == -1) {
        cout << "NO" << '\n';
        continue;
      }
      auto solve = [&](vi r) {
        vi ans;
        for (long long i = 0; i < r.size(); i += 2) {
          long long p = r[i] + 1;
          while (p + 1 < r[i + 1]) {
            ans.push_back(p);
            p += 2;
          }
          if (p == r[i + 1]) {
            for (long long j = r[i]; j < r[i + 1]; j += 2) {
              ans.push_back(j);
            }
          } else {
            for (long long j = r[i + 1] - 1; j >= r[i]; j -= 2) {
              ans.push_back(j);
            }
          }
        }
        return ans;
      };
      vi r;
      for (long long i = pos + 1; i < n; ++i) {
        if (a[i]) {
          r.push_back(i - pos);
        }
      }
      vi lp;
      for (long long i = pos - 1; i >= 0; --i) {
        if (a[i]) {
          lp.push_back(pos - i);
        }
      }
      for (auto& e : solve(r)) {
        ans.push_back(e + pos);
      }
      for (auto& e : solve(lp)) {
        ans.push_back(pos - e);
      }
      assert(ans.size() <= n);
      cout << "YES" << '\n' << ans.size() << '\n' << ans << '\n';
    }
  }
  