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>
  #pragma GCC optimize("O3")
  #pragma GCC target("tune=native")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  const int MAX = 1e6 + 5;
  const long long MOD = 1000000007;
  const long long MOD2 = 2010405347;
  const long long INF = 2e18;
  const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
  const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
  const double pi = acos(-1);
  const double EPS = 1e-9;
  const int block = 315;
  inline long long pw(long long x, long long y, long long md = MOD) {
    long long ret = 1;
    x %= md;
    while (y) {
      if (y & 1) ret = ret * x % md;
      x = x * x % md, y >>= 1;
    }
    return ret;
  }
  inline void add(int &a, int b, int md = MOD) {
    a = a + b >= md ? a + b - md : a + b;
  }
  long long tc, n, x[MAX], y[MAX], cnt, hit;
  vector<int> ans;
  void solve(int le, int ri) {
    cnt = x[le] + x[le + 1];
    int tmp = 0;
    for (int i = (le); i <= (ri - 2); ++i) {
      cnt += x[i + 2];
      if (cnt % 2 == 0 && cnt) {
        tmp = 0;
        for (int k = (0); k <= (2); ++k) tmp += x[i + k];
        if (tmp == 2) {
          ans.push_back(i);
          for (int k = (0); k <= (2); ++k) x[i + k] = 0;
          for (int j = (i); j >= (le); --j) {
            tmp = 0;
            for (int k = (0); k <= (2); ++k) tmp += x[j + k];
            if (tmp == 2) {
              ans.push_back(j);
              for (int k = (0); k <= (2); ++k) x[j + k] = 0;
            }
          }
          for (int j = (i); j <= (ri - 2); ++j) {
            tmp = 0;
            for (int k = (0); k <= (2); ++k) tmp += x[j + k];
            if (tmp == 2) {
              ans.push_back(j);
              for (int k = (0); k <= (2); ++k) x[j + k] = 0;
            }
          }
          return;
        }
      }
    }
    return;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> tc;
    for (int ntc = (1); ntc <= (tc); ++ntc) {
      cin >> n;
      cnt = 0;
      for (int i = (1); i <= (n); ++i) {
        cin >> x[i];
        y[i] = x[i];
        cnt += x[i];
      }
      if ((cnt & 1) || cnt == n) {
        cout << "NO\n";
        continue;
      }
      ans.clear();
      hit = cnt;
      cnt = 0;
      bool st = 0;
      for (int i = (1); i <= (n); ++i)
        cnt += x[i], st |= (cnt % 2 == 0) && (i & 1);
      cnt = 0;
      for (int i = (1); i <= (n); ++i) {
        cnt += x[i];
        if (cnt && (cnt % 2 == 0) && (i & 1)) {
          hit -= cnt;
          for (int j = (1); j <= (i - 2); ++j)
            if (x[j]) {
              cnt = 0;
              for (int k = (0); k <= (2); ++k) cnt += x[j + k];
              if (cnt == 1) {
                for (int k = (0); k <= (2); ++k) x[j + k] = 1;
                ans.push_back(j);
              }
            }
          solve(1, i);
          break;
        }
      }
      cnt = 0;
      for (int i = (n); i >= (1); --i) {
        cnt += x[i];
        if (cnt == hit && (i % 2 == n % 2)) {
          for (int j = (i); j <= (n - 2); ++j)
            if (x[j]) {
              cnt = 0;
              for (int k = (0); k <= (2); ++k) cnt += x[j + k];
              if (cnt == 1) {
                for (int k = (0); k <= (2); ++k) x[j + k] = 1;
                ans.push_back(j);
              }
            }
          solve(i, n);
          break;
        }
      }
      cnt = 0;
      for (int i = (1); i <= (n); ++i) {
        cnt += x[i];
      }
      if (cnt) {
        cout << "NO\n";
      } else {
        cout << "YES\n";
        assert(ans.size() <= n);
        cout << ans.size() << '\n';
        for (int i : ans) cout << i << " ";
        cout << '\n';
      }
    }
    return 0;
  }
  