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;
  long long mod = 1e9 + 7;
  long long po(long long x) { return ((x % mod) + mod) % mod; }
  long long SM(long long x, long long y) { return (po(x) + po(y)) % mod; }
  long long M(long long x, long long y) { return (po(x) * po(y)) % mod; }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long rand(long long x, long long y) {
    return uniform_int_distribution<long long>(x, y)(rng);
  }
  long long t = 1, n, i, j, k, x, y, z, l, r, mid, m, ans;
  long long a[400005], b[400005], c[400005], d[400005];
  map<long long, long long> mp;
  string s, st;
  vector<long long> v[300005], u;
  long long ff() {
    x = 0;
    u.clear();
    cin >> n;
    for (i = 0; i < n; i++) cin >> a[i], x += a[i];
    if (x % 2 || x == n) return cout << "NO", 0;
    if (x == 0) return cout << "YES\n0", 0;
    if (a[0]) {
      for (i = 0; i + 2 < n; i++)
        if (a[i] + a[i + 1] + a[i + 2] == 1) {
          a[i] = a[i + 1] = a[i + 2] = 1;
          u.push_back(i + 1);
        } else if (a[i] + a[i + 1] + a[i + 2] == 2 && i % 2 == 0) {
          a[i] = a[i + 1] = a[i + 2] = 0;
          u.push_back(i + 1);
          break;
        }
      if (i + 2 == n) return cout << "NO", 0;
      while (i >= 0) {
        if (a[i] + a[i + 1] + a[i + 2] == 2) {
          a[i] = a[i + 1] = a[i + 2] = 0;
          u.push_back(i + 1);
        }
        i--;
      }
    }
    for (i = 0; i + 2 < n; i++)
      if (a[i] + a[i + 1] + a[i + 2] == 2) {
        u.push_back(i + 1);
        a[i] = a[i + 1] = a[i + 2] = 0;
      } else if (a[i] && a[i + 1] + a[i + 2] == 0) {
        u.push_back(i + 1);
        u.push_back(i);
        swap(a[i], a[i + 2]);
      }
    cout << "YES";
    cout << "\n" << u.size() << "\n";
    for (auto i : u) cout << i << " ";
    return 0;
  }
  int main() {
    ios::sync_with_stdio(NULL);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t--) {
      ff();
      cout << "\n";
    }
  }
  