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("O3")
  using namespace std;
  unsigned seed = chrono::system_clock::now().time_since_epoch().count();
  mt19937 eng(seed);
  long long random2() { return (1ll << 31ll) * eng() + eng(); }
  long long n, m, k, q, T;
  const long long big = 1000000007;
  const long long big2 = 1000000009;
  const long long mod = 998244353;
  const long long MAXN = 400004;
  vector<int> A;
  vector<int> ANS;
  void mov(long long i) {
    ANS.push_back(i);
    long long x = (A[i] ^ A[i + 1] ^ A[i + 2]);
    for (int c1 = 0; c1 < (3); ++c1) {
      A[i + c1] = x;
    }
  }
  void even(long long i, long long j) {
    if (j == n - 1) {
      for (int c1 = i - 1; c1 < j; c1 += 2) {
        mov(c1);
      }
    } else {
      for (int c1 = j - 1; c1 >= i; c1 -= 2) {
        mov(c1);
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long a, b, c, d, e;
    cin >> T;
    for (int c4 = 0; c4 < (T); ++c4) {
      cin >> n;
      A.clear();
      ANS.clear();
      b = 0;
      for (int c1 = 0; c1 < (n); ++c1) {
        cin >> a;
        A.push_back(a);
        b += a;
      }
      if (b % 2 == 1 || b == n) {
        cout << "NO\n";
        continue;
      }
      long long start = -1;
      for (int c1 = 0; c1 < (n); ++c1) {
        if (A[c1] == 0) {
          if (start == -1) continue;
          int parity = (c1 - start) % 2;
          if (parity == 0) {
            even(start, c1 - 1);
            start = -1;
          } else {
            if (A[c1 + 1] == 1) {
              mov(c1 - 1);
              even(start, c1 - 2);
              start = -1;
            } else {
              mov(c1 - 1);
            }
          }
        } else {
          if (start == -1) {
            start = c1;
          }
        }
      }
      start = -1;
      for (int c1 = 0; c1 < (n); ++c1) {
        if (A[c1] == 1) {
          start = c1;
          break;
        }
      }
      if (start != -1) {
        if (start == 0) {
          cout << "NO\n";
          continue;
        } else {
          even(start, n - 1);
        }
      }
      cout << "YES\n";
      cout << (int)(ANS).size() << "\n";
      for (auto& y : ANS) {
        cout << y + 1 << " ";
      }
      cout << "\n";
    }
    return 0;
  }
  