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 inf = 1000000001;
  const int MOD = 1000000007;
  const int N = 200009;
  int a[N + 1];
  void doit(int pos, vector<int>& ans) {
    ans.push_back(pos);
    int t = a[pos] ^ a[pos + 1] ^ a[pos + 2];
    a[pos] = a[pos + 1] = a[pos + 2] = t;
  }
  void solve() {
    int n;
    scanf("%d", &n);
    vector<int> ans;
    int tot = 0;
    for (int i = (1); i <= (n); ++i) {
      scanf("%d", &a[i]);
      tot += a[i];
    }
    if (tot & 1) {
      cout << "NO" << endl;
      return;
    }
    int pr = 0;
    for (int i = (1); i <= (n); ++i) {
      if (!a[i]) {
        if (pr & 1) {
          int j = i + 1;
          while (j < n && !a[j]) j++;
          if (j == n + 1) {
            cout << "NO" << endl;
            return;
          }
          if ((j - i) % 2 == 0) {
            for (int f = j - 2; f >= i; f -= 2) doit(f, ans);
            pr += (j - i + 1);
            i = j;
            continue;
          }
          for (int f = j - 2; f > i; f -= 2) doit(f, ans);
          doit(i - 1, ans);
          for (int f = i - 3; f >= 1; f -= 2) {
            if (!a[f]) break;
            doit(f, ans);
          }
          for (int f = i + 1; f < j; f += 2) doit(f, ans);
          pr = 0;
        } else {
          for (int f = i - 2; f >= 1; f -= 2) {
            if (!a[f]) break;
            doit(f, ans);
          }
          pr = 0;
        }
      } else {
        pr++;
      }
    }
    if (pr & 1 || pr == n) {
      cout << "NO" << endl;
      return;
    }
    for (int f = n - pr; f < n; f += 2) doit(f, ans);
    cout << "YES" << endl;
    cout << (int)((ans).size()) << endl;
    for (int i = (0); i <= ((int)((ans).size()) - 1); ++i) {
      printf("%d ", ans[i]);
    }
    cout << endl;
  }
  int main(int argc, const char* argv[]) {
    int tests;
    scanf("%d\n", &tests);
    for (int i = (0); i <= (tests - 1); ++i) {
      solve();
    }
    return 0;
  }
  