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;
  const int N = 2e5 + 5;
  int a[N];
  vector<int> ans;
  void work(int l, int r) {
    if (a[l]) {
      assert(a[r] == 0);
      int t = ans.size();
      reverse(a + l, a + r + 1);
      work(l, r);
      for (int i = t; i < ans.size(); ++i) {
        int p = ans[i] - l + 1;
        p = p + 2;
        p = r - p + 1;
        ans[i] = l + p - 1;
      }
      return;
    }
    while (l < r) {
      if (a[l + 1] == 0) {
        ++l;
        continue;
      }
      assert(l + 1 != r);
      if (a[l + 2]) {
        ans.push_back(l);
        a[l + 1] = a[l + 2] = 0;
        continue;
      }
      assert(l + 2 != r);
      if (a[l + 3]) {
        ans.push_back(l + 1);
        a[l + 1] = a[l + 3] = 0;
        continue;
      }
      ans.push_back(l + 1);
      a[l + 2] = a[l + 3] = 1;
    }
  }
  int main() {
    int tt;
    cin >> tt;
    while (tt--) {
      int n;
      cin >> n;
      int sum = 0;
      for (int i = 1; i <= n; ++i) {
        scanf("%d", a + i);
        sum ^= a[i];
      }
      if (sum % 2) {
        puts("NO");
        continue;
      }
      ans.clear();
      if (a[1] == 0 || a[n] == 0) {
        work(1, n);
      } else {
        int p_sum = 0, i = 1;
        for (i = 1; i <= n; ++i)
          if (a[i])
            p_sum ^= 1;
          else if (p_sum == 0) {
            work(1, i);
            work(i, n);
            break;
          }
        if (i > n) {
          for (i = 1; i <= n; ++i)
            if (a[i] == 0) {
              int j = i;
              while (j <= n && a[j] == 0) ++j;
              --j;
              if ((j - i + 1) % 2) {
                while (j > i) {
                  ans.push_back(j - 1);
                  a[j - 1] = a[j] = 1;
                  j -= 2;
                }
                ans.push_back(i - 1);
                a[i - 1] = a[i + 1] = 0;
                work(1, i);
                work(i, n);
                break;
              }
              i = j + 1;
            }
          if (i > n) {
            puts("NO");
            continue;
          }
        }
      }
      puts("YES");
      cout << ans.size() << endl;
      for (auto x : ans) printf("%d ", x);
      puts("");
    }
  }
  