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;
  mt19937 rng((long long)chrono::steady_clock::now().time_since_epoch().count());
  const long long mod = 1e9 + 7;
  long long n, a[200010];
  void op_ida(long long Ini, vector<long long> &ans) {
    for (long long ini = Ini; ini <= n - 2; ini++) {
      long long meio = ini + 1, fim = ini + 2;
      if (a[meio] == 0) continue;
      if (a[fim] == 1) {
        ans.push_back(ini);
        a[meio] = a[fim] = 0;
        continue;
      }
      if (meio <= n - 2) {
        if (a[fim + 1] == 1) {
          ans.push_back(meio);
          a[meio] = a[fim + 1] = 0;
        } else {
          ans.push_back(meio);
          ans.push_back(ini);
          a[meio] = a[fim] = 0;
          a[fim + 1] = 1;
        }
        continue;
      }
      cout << "NO\n";
      exit(0);
    }
  }
  void op_volta(long long Ini, vector<long long> &ans) {
    for (long long ini = Ini; ini >= 3; ini--) {
      long long meio = ini - 1, fim = ini - 2;
      if (a[meio] == 0) continue;
      if (a[fim] == 1) {
        ans.push_back(fim);
        a[meio] = a[fim] = 0;
        continue;
      }
      if (meio >= 3) {
        if (a[fim - 1] == 1) {
          ans.push_back(fim - 1);
          a[meio] = a[fim - 1] = 0;
        } else {
          ans.push_back(fim - 1);
          ans.push_back(fim);
          a[meio] = a[fim] = 0;
          a[fim - 1] = 1;
        }
        continue;
      }
      cout << "NO\n";
      exit(0);
    }
  }
  void solve() {
    cin >> n;
    a[n + 1] = a[0] = 0;
    long long tot = 0;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
      tot += a[i];
    }
    if (tot & 1 or tot == n) {
      cout << "NO\n";
      return;
    }
    if (a[1] == 0) {
      vector<long long> ans;
      op_ida(1, ans);
      cout << "YES\n";
      cout << ans.size() << "\n";
      for (auto u : ans) cout << u << " ";
      cout << "\n";
      return;
    }
    if (a[n] == 0) {
      vector<long long> ans;
      op_volta(n, ans);
      cout << "YES\n";
      cout << ans.size() << "\n";
      for (auto u : ans) cout << u << " ";
      cout << "\n";
      return;
    }
    long long pref = 0;
    for (long long i = 1; i <= n; i++) {
      pref += a[i];
      if (a[i] == 1) continue;
      if (pref % 2 == 0) {
        vector<long long> ans;
        op_ida(i, ans);
        op_volta(i, ans);
        cout << "YES\n";
        cout << ans.size() << "\n";
        for (auto u : ans) cout << u << " ";
        cout << "\n";
        return;
      } else {
        long long j = i;
        while (j <= n and a[j] == 0) j++;
        j--;
        if ((j - i + 1) & 1) {
          vector<long long> ans;
          while (j != i) {
            a[j] = a[j - 1] = 1;
            ans.push_back(j - 1);
            j -= 2;
          }
          ans.push_back(i - 1);
          a[i + 1] = 0;
          a[i - 1] = 0;
          op_ida(i, ans);
          op_volta(i, ans);
          cout << "YES\n";
          cout << ans.size() << "\n";
          for (auto u : ans) cout << u << " ";
          cout << "\n";
          return;
        }
        i = j;
      }
    }
    cout << "NO\n";
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long t;
    cin >> t;
    for (long long i = 1; i <= t; i++) solve();
  }
  