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 = 200005;
  int n, a[N];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) {
      cin >> n;
      int cnt1 = 0, cnt0 = 0;
      for (int i = 1; i <= n; i++) {
        cin >> a[i];
        if (a[i])
          cnt1++;
        else
          cnt0++;
      }
      if (cnt1 % 2 != 0 || !cnt0)
        cout << "NO" << endl;
      else if (!cnt1)
        cout << "YES\n0" << endl;
      else {
        vector<int> ans;
        int p = 1, fucked = 0;
        while (1) {
          while (p <= n && a[p]) p++;
          if (p > n) {
            fucked = 1;
            break;
          }
          if (p % 2 == 1)
            break;
          else {
            bool ok = 1;
            while (!a[p + 1]) {
              ans.push_back(p - 1);
              a[p] = a[p + 1] = 1;
              p += 2;
              if (a[p]) {
                ok = 0;
                break;
              }
            }
            if (ok) {
              ans.push_back(p - 1);
              a[p - 1] = a[p + 1] = 0;
              p--;
              break;
            }
          }
        }
        if (fucked) {
          cout << "NO" << endl;
          continue;
        }
        int q = p;
        while (p - 2 > 0) {
          ans.push_back(p - 2);
          a[p - 2] = a[p - 1] = 0;
          p -= 2;
        }
        while (1) {
          while (q <= n && !a[q]) q++;
          if (q > n) break;
          p = q;
          while (1) {
            while (p <= n && a[p]) p++;
            if ((p - q) % 2 == 0)
              break;
            else {
              bool ok = 1;
              while (!a[p + 1]) {
                ans.push_back(p - 1);
                a[p] = a[p + 1] = 1;
                p += 2;
                if (a[p]) {
                  ok = 0;
                  break;
                }
              }
              if (ok) {
                ans.push_back(p - 1);
                a[p - 1] = a[p + 1] = 0;
                p--;
                break;
              }
            }
          }
          while (q < p) {
            ans.push_back(q - 1);
            q += 2;
          }
        }
        cout << "YES\n" << ans.size() << endl;
        for (int i : ans) cout << i << ' ';
        cout << endl;
      }
    }
  }
  