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;
  using ll = long long;
  using ull = unsigned long long;
  using ld = long double;
  using db = double;
  using str = string;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using pd = pair<db, db>;
  using vi = vector<int>;
  using vb = vector<bool>;
  using vl = vector<ll>;
  using vd = vector<db>;
  using vs = vector<str>;
  using vpi = vector<pi>;
  using vpl = vector<pl>;
  using vpd = vector<pd>;
  template <class T>
  using V = vector<T>;
  template <class T, size_t SZ>
  using AR = array<T, SZ>;
  template <class T>
  using PR = pair<T, T>;
  template <class T>
  int lwb(V<T>& a, const T& b) {
    return int(lower_bound(begin(a), end(a), b) - begin(a));
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int tt;
    cin >> tt;
    for (int ti = 0; ti < tt; ti++) {
      int n;
      cin >> n;
      vector<int> a(n);
      vector<int> c(2);
      int first = -1;
      int second = -1;
      for (int i = (0); i < (n); ++i) {
        cin >> a[i];
        c[a[i]]++;
        if (a[i] == 0 && first == -1 && c[1] % 2 == 0) first = i;
      }
      for (int i = (0); i < (n); ++i) {
        if (i > 0 && i < n - 1 && a[i] == 0 && a[i - 1] == 1 && a[i + 1] == 1)
          second = i;
      }
      if (c[1] == 0) {
        cout << "YES" << endl << 0 << endl;
        continue;
      }
      if (c[1] % 2 == 1 || c[0] == 0) {
        cout << "NO" << endl;
        continue;
      }
      vector<int> ans;
      if (first == -1) {
        if (second == -1) {
          int last = -1;
          for (int i = (0); i < (n); ++i) {
            if (a[i] == 0) {
              if (last == -1) last = i;
            } else {
              if (last != -1) {
                if (i - last >= 3 && (i - last) % 2 == 1) {
                  for (int j = last; j < i - 2; j += 2) {
                    ans.push_back(j - 1);
                    a[j] = a[j + 1] = 1;
                  }
                  second = i - 1;
                  break;
                }
              }
              last = -1;
            }
          }
        }
        if (second != -1) {
          ans.push_back(second - 1);
          a[second] = 0;
          a[second - 1] = a[second + 1] = 0;
          first = second - 1;
        }
        if (first == -1) {
          cout << "NO" << endl;
          continue;
        }
      }
      if (first > 0) {
        int i = first - 1;
        while (i >= 0) {
          if (a[i] == 1) {
            if (i - 1 >= 0 && a[i - 1] == 1) {
              ans.push_back(i - 1);
              a[i] = a[i - 1] = 0;
              i = i - 2;
            } else {
              if (i - 2 >= 0 && a[i - 2] == 1) {
                ans.push_back(i - 2);
                a[i] = a[i - 2] = 0;
                i = i - 3;
              } else {
                ans.push_back(i - 2);
                ans.push_back(i - 1);
                a[i] = 0;
                i = i - 2;
                a[i] = 1;
              }
            }
          } else {
            i = i - 1;
          }
        }
      }
      if (first < n - 1) {
        int i = first + 1;
        while (i < n) {
          if (a[i] == 1) {
            if (i + 1 < n & a[i + 1] == 1) {
              ans.push_back(i - 1);
              a[i] = a[i + 1] = 0;
              i += 2;
            } else {
              if (i + 2 < n && a[i + 2] == 1) {
                ans.push_back(i);
                a[i] = a[i + 2] = 0;
                i = i + 3;
              } else {
                ans.push_back(i);
                ans.push_back(i - 1);
                a[i] = 0;
                i = i + 2;
                a[i] = 1;
              }
            }
          } else {
            i = i + 1;
          }
        }
      }
      cout << "YES" << endl;
      cout << ans.size() << endl;
      if (ans.size() > 0) {
        for (int i = (0); i < (ans.size()); ++i) {
          if (i > 0) cout << " ";
          cout << ans[i] + 1;
        }
        cout << endl;
      }
    }
  }
  