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], ans[N], n, tot;
  bool dfs(int x) {
    if (x > n - 2) return 0;
    if (a[x] + a[x + 1] + a[x + 2] == 2) {
      ans[++tot] = x;
      a[x] = a[x + 1] = a[x + 2] = a[x] ^ a[x + 1] ^ a[x + 2];
      return 1;
    }
    if (!a[x]) return 1;
    if (!a[x + 2]) {
      ans[++tot] = x;
      a[x] = a[x + 1] = a[x + 2] = a[x] ^ a[x + 1] ^ a[x + 2];
    }
    if (dfs(x + 2)) {
      ans[++tot] = x;
      a[x] = a[x + 1] = a[x + 2] = a[x] ^ a[x + 1] ^ a[x + 2];
      return 1;
    }
    return 0;
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      cin >> n;
      int s = 0;
      for (int i = 1; i <= n; i++) {
        cin >> a[i];
        s += a[i];
      }
      if (s & 1) {
        cout << "NO" << endl;
        continue;
      }
      tot = 0;
      int flag = 1;
      for (int i = 1; i <= n - 2; i++) {
        if (!dfs(i)) {
          if (i == 1 || !dfs(i - 1)) {
            flag = 0;
            cout << "NO" << endl;
            goto aaa;
          }
        }
      }
      if (flag) {
        cout << "YES" << endl;
        cout << tot << endl;
        for (int i = 1; i <= tot; i++) cout << ans[i] << " ";
        if (tot) cout << endl;
      }
    aaa:;
    }
  }
  