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;
  inline int read() {
    char c = getchar();
    int x = 0;
    bool f = 0;
    for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
    for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    if (f) x = -x;
    return x;
  }
  struct modint {
    int x;
    modint(int o = 0) { x = o; }
    modint &operator=(int o) { return x = o, *this; }
    modint &operator+=(modint o) {
      return x = x + o.x >= 1000000007 ? x + o.x - 1000000007 : x + o.x, *this;
    }
    modint &operator-=(modint o) {
      return x = x - o.x < 0 ? x - o.x + 1000000007 : x - o.x, *this;
    }
    modint &operator*=(modint o) { return x = 1ll * x * o.x % 1000000007, *this; }
    modint &operator^=(int b) {
      modint a = *this, c = 1;
      for (; b; b >>= 1, a *= a)
        if (b & 1) c *= a;
      return x = c.x, *this;
    }
    modint &operator/=(modint o) { return *this *= o ^= 1000000007 - 2; }
    modint &operator+=(int o) {
      return x = x + o >= 1000000007 ? x + o - 1000000007 : x + o, *this;
    }
    modint &operator-=(int o) {
      return x = x - o < 0 ? x - o + 1000000007 : x - o, *this;
    }
    modint &operator*=(int o) { return x = 1ll * x * o % 1000000007, *this; }
    modint &operator/=(int o) { return *this *= ((modint(o)) ^= 1000000007 - 2); }
    template <class I>
    friend modint operator+(modint a, I b) {
      return a += b;
    }
    template <class I>
    friend modint operator-(modint a, I b) {
      return a -= b;
    }
    template <class I>
    friend modint operator*(modint a, I b) {
      return a *= b;
    }
    template <class I>
    friend modint operator/(modint a, I b) {
      return a /= b;
    }
    friend modint operator^(modint a, int b) { return a ^= b; }
    friend bool operator==(modint a, int b) { return a.x == b; }
    friend bool operator!=(modint a, int b) { return a.x != b; }
    bool operator!() { return !x; }
    modint operator-() { return x ? 1000000007 - x : 0; }
  };
  int n, a[4000005], sum[4000005];
  vector<int> res;
  void work(int l, int r) {
    for (int i = r - 2; i >= l; i -= 2) res.push_back(i);
    for (int i = l + 2; i <= r - 2; i += 2) res.push_back(i);
  }
  void work() {
    n = read();
    res.clear();
    for (register int i = (1); i <= (n); ++i)
      a[i] = read(), sum[i] = sum[i - 1] ^ a[i];
    if (sum[n]) return puts("NO"), void();
    if (n & 1)
      work(1, n);
    else {
      bool ok = 0;
      for (register int i = (1); i <= (n); ++i)
        if (!sum[i] && (i & 1)) {
          ok = 1;
          work(1, i), work(i + 1, n);
          break;
        }
      if (!ok) return puts("NO"), void();
    }
    puts("YES");
    cout << res.size() << endl;
    for (auto v : res) cout << v << " ";
    puts("");
  }
  signed main() {
    int T = read();
    while (T--) work();
    return 0;
  }
  