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>
  #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
  using namespace std;
  template <class T>
  using rque = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  bool chmin(T &a, const T &b) {
    if (b < a) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  bool chmax(T &a, const T &b) {
    if (b > a) {
      a = b;
      return 1;
    }
    return 0;
  }
  long long gcd(long long a, long long b) {
    if (a == 0) return b;
    if (b == 0) return a;
    long long cnt = a % b;
    while (cnt != 0) {
      a = b;
      b = cnt;
      cnt = a % b;
    }
    return b;
  }
  long long extGCD(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    long long d = extGCD(b, a % b, y, x);
    y -= a / b * x;
    return d;
  }
  struct UnionFind {
    vector<long long> data;
    int num;
    UnionFind(int sz) {
      data.assign(sz, -1);
      num = sz;
    }
    bool unite(int x, int y) {
      x = find(x), y = find(y);
      if (x == y) return (false);
      if (data[x] > data[y]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
      num--;
      return (true);
    }
    int find(int k) {
      if (data[k] < 0) return (k);
      return (data[k] = find(data[k]));
    }
    long long size(int k) { return (-data[find(k)]); }
    bool same(int x, int y) { return find(x) == find(y); }
  };
  template <int mod>
  struct ModInt {
    int x;
    ModInt() : x(0) {}
    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
    ModInt &operator+=(const ModInt &p) {
      if ((x += p.x) >= mod) x -= mod;
      return *this;
    }
    ModInt &operator-=(const ModInt &p) {
      if ((x += mod - p.x) >= mod) x -= mod;
      return *this;
    }
    ModInt &operator*=(const ModInt &p) {
      x = (int)(1LL * x * p.x % mod);
      return *this;
    }
    ModInt &operator/=(const ModInt &p) {
      *this *= p.inverse();
      return *this;
    }
    ModInt operator-() const { return ModInt(-x); }
    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
    bool operator==(const ModInt &p) const { return x == p.x; }
    bool operator!=(const ModInt &p) const { return x != p.x; }
    ModInt inverse() const {
      int a = x, b = mod, u = 1, v = 0, t;
      while (b > 0) {
        t = a / b;
        swap(a -= t * b, b);
        swap(u -= t * v, v);
      }
      return ModInt(u);
    }
    ModInt pow(int64_t n) const {
      ModInt ret(1), mul(x);
      while (n > 0) {
        if (n & 1) ret *= mul;
        mul *= mul;
        n >>= 1;
      }
      return ret;
    }
    friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }
    friend istream &operator>>(istream &is, ModInt &a) {
      int64_t t;
      is >> t;
      a = ModInt<mod>(t);
      return (is);
    }
    static int get_mod() { return mod; }
  };
  long long mpow2(long long x, long long n, long long mod) {
    long long ans = 1;
    while (n != 0) {
      if (n & 1) ans = ans * x % mod;
      x = x * x % mod;
      n = n >> 1;
    }
    return ans;
  }
  constexpr int mod = 1000000007;
  using mint = ModInt<mod>;
  mint mpow(mint x, long long n) {
    mint ans = 1;
    while (n != 0) {
      if (n & 1) ans *= x;
      x *= x;
      n = n >> 1;
    }
    return ans;
  }
  int main() {
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    cout << fixed << setprecision(15);
    int T;
    cin >> T;
    while (T--) {
      long long n;
      cin >> n;
      long long a[n];
      for (long long i = 0; i < (n); i++) cin >> a[i];
      vector<long long> ans;
      int f = 0, o = 0;
      for (long long i = 0; i < (n); i++) {
        if (a[i]) {
          o ^= 1;
          f = 1;
        } else {
          if (f) {
            if (o) {
              if (i < n - 1) {
                if (a[i + 1]) {
                  ans.emplace_back(i);
                  a[i - 1] = 0;
                  a[i + 1] = 0;
                  long long idx = i - 1;
                  while (1) {
                    if (idx < 2) break;
                    if (a[idx - 1] && a[idx - 2]) {
                      ans.emplace_back(idx - 1);
                      a[idx - 2] = 0;
                      a[idx - 1] = 0;
                      idx -= 2;
                    } else
                      break;
                  }
                  f = 0;
                  o = 0;
                } else {
                  ans.emplace_back(i);
                  a[i] = 1;
                  a[i + 1] = 1;
                  o = 0;
                }
              }
            } else {
              long long idx = i;
              while (1) {
                if (idx < 2) break;
                if (a[idx - 1] && a[idx - 2]) {
                  ans.emplace_back(idx - 1);
                  a[idx - 2] = 0;
                  a[idx - 1] = 0;
                  idx -= 2;
                } else
                  break;
              }
              f = 0;
              o = 0;
            }
          }
        }
      }
      if (f) {
        long long idx = n - 1;
        while (idx >= 0) {
          if (a[idx])
            idx--;
          else
            break;
        }
        if (idx >= 0) {
          while (1) {
            if (idx + 2 >= n) break;
            if (a[idx + 1] && a[idx + 2]) {
              ans.emplace_back(idx + 1);
              a[idx + 2] = 0;
              a[idx + 1] = 0;
              idx += 2;
            } else
              break;
          }
        }
      }
      int flag = 1;
      for (long long i = 0; i < (n); i++)
        if (a[i]) flag = 0;
      if (flag) {
        cout << "YES" << endl;
        cout << ans.size() << endl;
        for (long long i = 0; i < (ans.size()); i++)
          cout << ans[i] << (i + 1 == ans.size() ? '\n' : ' ');
      } else
        cout << "NO" << endl;
    }
  }
  