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;
  int a[200005];
  int p[200005];
  int n, m;
  pair<int, int> t[200005];
  int len;
  vector<int> ans;
  void solve(int x) {
    ans.emplace_back(x);
    int val = a[x] ^ a[x + 1] ^ a[x + 2];
    a[x] = a[x + 1] = a[x + 2] = val;
  }
  void print() {
    cout << "YES\n";
    cout << ans.size() << '\n';
    if (ans.empty()) return;
    for (auto u : ans) cout << u << ' ';
    cout << '\n';
  }
  void solve() {
    cin >> n;
    ans.clear();
    m = len = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (int i = 1; i <= n; i++)
      if (a[i]) p[++m] = i;
    if ((m & 1) || n == m) {
      cout << "NO\n";
      return;
    }
    int l = 1;
    p[m + 1] = 1e9;
    for (int i = 2; i <= m + 1; i++) {
      if (p[i] > p[i - 1] + 1) {
        t[++len] = make_pair(p[l], p[i - 1]);
        l = i;
      }
    }
    for (int i = 1; i <= len; i++)
      if ((t[i].second - t[i].first + 1) & 1) {
        int pos = t[i].second;
        while (t[i + 1].first - pos + 1 > 3) {
          solve(pos);
          pos += 2;
        }
        if (pos == t[i + 1].first - 1) {
          t[i + 1].first = t[i].first;
          continue;
        }
        while (pos >= t[i].first) {
          solve(pos);
          pos -= 2;
        }
        t[i + 1].first++;
      } else {
        if (t[i].first != 1) {
          for (int j = t[i].first; j <= t[i].second; j += 2) {
            solve(j - 1);
          }
        } else if (t[i].second != n) {
          for (int j = t[i].second; j >= t[i].first; j -= 2) {
            solve(j - 1);
          }
        }
      }
    for (int i = 1; i <= n; i++)
      if (a[i]) {
        cout << "NO\n";
        return;
      }
    print();
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) solve();
    return 0;
  }
  