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;
  #pragma GCC optimize("Ofast,unroll-loops")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2,fma,tune=native")
  template <typename T>
  void print(T&& x) {
    cout << x << "\n";
  }
  template <typename T, typename... S>
  void print(T&& x, S&&... y) {
    cout << x << ' ';
    print(y...);
  }
  const long long Mod = 1000000007, Mod2 = 998244353;
  const long long MOD = Mod;
  const long long maxn = 200005;
  long long n, a[maxn];
  vector<long long> res;
  void op(long long x) {
    long long k = a[x] ^ a[x + 1] ^ a[x + 2];
    a[x] = a[x + 1] = a[x + 2] = k;
    res.push_back(x);
  }
  void orzck() {
    cin >> n;
    for (long long i = (0); i < ((long long)(n)); i++) cin >> a[i];
    if (accumulate(a, a + n, 0ll) & 1) {
      print("NO");
      return;
    }
    res.clear();
    826;
    long long cnt = 0;
    long long x = -1;
    for (long long i = (0); i < ((long long)(n)); i++) {
      if (a[i])
        cnt++;
      else if (cnt % 2 == 0) {
        x = i;
        break;
      }
    }
    if (x == -1) {
      for (long long i = (0); i < ((long long)(n - 2)); i++) {
        if (a[i + 1] == 0) {
          if (a[i + 2]) {
            op(i);
            x = i;
            break;
          }
          op(i);
        }
      }
    }
    if (x == -1) {
      print("NO");
      return;
    }
    long long l = x;
    while (l < n - 1) {
      if (a[l + 1] == 0) {
        l++;
        continue;
      }
      if (a[l + 2] == 1) {
        op(l);
        l += 2;
        continue;
      }
      if (a[l + 3] == 1) {
        op(l + 1);
        l += 3;
        continue;
      }
      op(l + 1);
      op(l);
      l += 2;
    }
    long long r = x;
    while (r) {
      if (a[r - 1] == 0) {
        r--;
        continue;
      }
      if (a[r - 2] == 1) {
        op(r - 2);
        r -= 2;
        continue;
      }
      if (a[r - 3] == 1) {
        op(r - 3);
        r -= 3;
        continue;
      }
      op(r - 3);
      op(r - 2);
      r -= 2;
    }
    for (long long i = (0); i < ((long long)(n)); i++) 826;
    print("YES");
    print(((long long)res.size()));
    for (auto i : res) cout << (i + 1) << ' ';
    cout << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    cout << fixed << setprecision(15);
    long long t;
    cin >> t;
    while (t--) orzck();
    return 0;
  }
  