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 + 10;
  int n, a[N], cnt;
  void solve() {
    bool flag = false;
    cnt = 0;
    cin >> n;
    vector<int> ones, ans;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (a[i]) {
        ones.push_back(i);
        cnt++;
      }
    }
    if (cnt & 1) {
      cout << "NO" << endl;
      return;
    }
    for (int i = 1; i < cnt; i += 2) {
      int firstIndex = ones[i - 1], secondIndex = ones[i];
      vector<int> curr;
      for (int j = firstIndex; j < secondIndex; j += 2) {
        if (j + 2 <= secondIndex) {
          int x = a[j] ^ a[j + 1] ^ a[j + 2];
          a[j] = x, a[j + 1] = x, a[j + 2] = x;
          curr.push_back(j);
          ans.push_back(j);
        }
      }
      if ((secondIndex - firstIndex - 1) & 1) {
        flag = true;
        for (int j = (int)curr.size() - 2; j >= 0; j--) {
          ans.push_back(curr[j]);
          int x = a[curr[j]] ^ a[curr[j] + 1] ^ a[curr[j] + 2];
          a[curr[j]] = x, a[curr[j] + 1] = x, a[curr[j] + 2] = x;
        }
      }
    }
    if (!flag) {
      for (int i = 1; i <= n; i++) {
        if (!a[i]) {
          flag = true;
          break;
        }
      }
      if (!flag) {
        cout << "NO" << endl;
        return;
      }
    }
    for (int i = 1; i <= n; i++) {
      int firstIndex = i, secondIndex = i;
      if (!a[i]) continue;
      while (secondIndex + 1 <= n && a[secondIndex + 1] == 1) {
        secondIndex++;
      }
      if (firstIndex == secondIndex) {
        continue;
      }
      if (firstIndex == 1) {
        for (int j = secondIndex - 1; j >= firstIndex; j -= 2) {
          ans.push_back(j);
        }
      } else {
        for (int j = firstIndex - 1; j < secondIndex; j += 2) {
          ans.push_back(j);
        }
      }
      i = secondIndex;
    }
    cout << "YES" << endl;
    cout << ans.size() << endl;
    for (auto i : ans) {
      cout << i << " ";
    }
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int T;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  