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>
  int main() {
    int T;
    std::cin >> T;
    while (T--) {
      int n;
      std::cin >> n;
      std::vector<int> a(n);
      for (int j = 0; j < n; ++j) std::cin >> a[j];
      std::vector<int> p;
      for (int j = 0; j < n; ++j) {
        if (a[j] == 1) p.push_back(j);
      }
      int sz = p.size();
      if (sz % 2 == 1) {
        std::cout << "NO" << std::endl;
      } else {
        std::vector<int> b;
        for (int j = 0; j < sz; j += 2) {
          int d = p[j + 1] - p[j];
          if (d % 2 == 0) {
            for (int k = p[j]; k < p[j + 1] - 2; k += 2) {
              b.push_back(k);
            }
            b.push_back(p[j + 1] - 2);
            for (int k = p[j + 1] - 4; k >= p[j]; k -= 2) {
              b.push_back(k);
            }
            a[p[j]] = a[p[j + 1]] = 0;
          } else {
            for (int k = p[j]; k <= p[j + 1] - 3; k += 2) {
              b.push_back(k), a[k + 1] = a[k + 2] = 1;
            }
          }
        }
        for (int j = 0; j < n - 2; ++j) {
          if (a[j] == 0 && a[j + 1] == 1 && a[j + 2] == 1) {
            b.push_back(j);
            a[j + 1] = a[j + 2] = 0;
          }
        }
        for (int j = n - 3; j >= 0; j--) {
          if (a[j] == 1 && a[j + 1] == 1 && a[j + 2] == 0) {
            b.push_back(j);
            a[j] = a[j + 1] = 0;
          }
        }
        bool ok = true;
        for (int j = 0; j < n; ++j) {
          if (a[j] == 1) ok = false;
        }
        if (!ok) {
          std::cout << "NO" << std::endl;
        } else {
          int k = b.size();
          std::cout << "YES" << std::endl;
          std::cout << k << std::endl;
          for (int j = 0; j < k; ++j) {
            std::cout << b[j] + 1 << " \n"[j == k - 1];
          }
        }
      }
    }
  }
  