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 ull = unsigned long long;
  using uint = unsigned;
  inline ll popcnt(ull a) { return __builtin_popcountll(a); }
  ll intpow(ll a, ll b) {
    ll ans = 1;
    while (b) {
      if (b & 1) ans *= a;
      a *= a;
      b /= 2;
    }
    return ans;
  }
  ll intpow(ll a, ll b, ll m) {
    ll ans = 1;
    while (b) {
      if (b & 1) (ans *= a) %= m;
      (a *= a) %= m;
      b /= 2;
    }
    return ans;
  }
  template <class T>
  inline bool chmax(T& a, T b) {
    if (a < b) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  inline bool chmin(T& a, T b) {
    if (a > b) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  using vc = vector<T>;
  template <class T>
  using vvc = vector<vc<T>>;
  template <class T>
  using vvvc = vector<vvc<T>>;
  template <class T>
  using vvvvc = vector<vvvc<T>>;
  template <class T>
  using pq = priority_queue<T>;
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  void scan(T& a) {
    cin >> a;
  }
  template <class T>
  void scan(vector<T>& a) {
    for (auto&& i : a) scan(i);
  }
  void in() {}
  template <class Head, class... Tail>
  void in(Head& head, Tail&... tail) {
    scan(head);
    in(tail...);
  }
  void print() { cout << ' '; }
  template <class T>
  void print(const T& a) {
    cout << a;
  }
  template <class T>
  void print(const vector<T>& a) {
    if (a.empty()) return;
    print(a[0]);
    for (auto i = a.begin(); ++i != a.end();) {
      cout << ' ';
      print(*i);
    }
  }
  int out() {
    cout << '\n';
    return 0;
  }
  template <class T>
  int out(const T& t) {
    print(t);
    cout << '\n';
    return 0;
  }
  template <class Head, class... Tail>
  int out(const Head& head, const Tail&... tail) {
    print(head);
    cout << ' ';
    out(tail...);
    return 0;
  }
  void solve() {
    long long n;
    in(n);
    vector<long long> a(n);
    in(a);
    vc<long long> ans;
    auto f = [&](long long i) -> void {
      assert(i >= 0 and i < n);
      ans.push_back(i);
      long long x = (a[i] ^ a[i + 1] ^ a[i + 2]);
      a[i] = x;
      a[i + 1] = x;
      a[i + 2] = x;
    };
    long long zcnt = 0, ocnt = 0;
    for (long long i = 0; i < n; ++i) {
      if (a[i] == 0)
        ++zcnt;
      else
        ++ocnt;
    }
    if (ocnt % 2 == 1) {
      out("NO");
      return;
    };
    long long cnt = 0;
    for (long long i = 0; i < n; ++i) {
      if (a[i] == 0) {
        if (cnt % 2 == 0) {
          cnt = 0;
        } else {
          if (i + 1 < n) {
            f(i - 1);
            if (a[i] == 0) {
              cnt = 0;
            } else {
              ++cnt;
            }
          }
        }
      } else {
        ++cnt;
      }
    };
    for (long long i = 0; i < n; ++i) {
      if (i - 1 >= 0 and a[i - 1] == 1 and a[i] == 0) {
        for (long long j = i - 2; j >= 0 and a[j] == 1 and a[j + 1] == 1;
             j -= 2) {
          f(j);
        }
      }
      if (i + 1 < n and a[i] == 0 and a[i + 1] == 1) {
        for (long long j = i; j + 2 < n and a[j + 2] == 1 and a[j + 1] == 1;
             j += 2) {
          f(j);
        }
      }
    };
    for (long long i = 0; i < n; ++i) {
      if (a[i] == 1) {
        out("NO");
        return;
      }
    }
    out("YES");
    for (long long& x : ans) ++x;
    out((long long)(ans).size());
    out(ans);
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    long long t;
    in(t);
    for (long long i = 0; i < t; ++i) solve();
    return 0;
  }
  