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 rnd(chrono::steady_clock::now().time_since_epoch().count());
  mt19937 rnf(2106);
  const int N = 200005;
  int n;
  int a[N];
  int p[N], s[N];
  vector<int> ans;
  void ubd(int i) {
    assert(1 <= i && i <= n - 2);
    ans.push_back(i);
    int x = (a[i] ^ a[i + 1] ^ a[i + 2]);
    a[i] = a[i + 1] = a[i + 2] = x;
  }
  bool solv() {
    cin >> n;
    for (int i = 0; i <= n + 1; ++i) {
      p[i] = s[i] = 0;
    }
    ans.clear();
    for (int i = 1; i <= n; ++i) cin >> a[i];
    bool z = false;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == 0) {
        z = true;
        break;
      }
    }
    if (!z) {
      cout << "NO\n";
      return false;
    }
    for (int i = 1; i <= n; ++i) p[i] = p[i - 1] + a[i];
    for (int i = n; i >= 1; --i) s[i] = s[i + 1] + a[i];
    if (p[n] % 2 == 1) {
      cout << "NO\n";
      return false;
    }
    int u = -1;
    for (int i = 1; i <= n; ++i) {
      if (a[i] == 0 && p[i] % 2 == 0 && s[i] % 2 == 0) {
        u = i;
        break;
      }
    }
    if (u == -1) {
      int l, r;
      bool f = false;
      for (int i = 1; i <= n; ++i) {
        if (a[i] == 0) {
          if (i == 0 || a[i - 1] == 1) {
            l = i;
          }
          r = i;
          if (i == n || a[i + 1] == 1) {
            if ((r - l + 1) % 2 == 1) {
              f = true;
              break;
            }
          }
        }
      }
      if (!f) {
        cout << "NO\n";
        return false;
      }
      for (int i = l - 1; i <= n; ++i) {
        if (a[i] == 1 && a[i + 1] == 0 && a[i + 2] == 1) {
          ubd(i);
          u = i;
          break;
        } else if (a[i] == 1 && a[i + 1] == 0 && a[i + 2] == 0) {
          ubd(i);
        }
      }
    }
    if (u != -1) {
      for (int i = u; i <= n; ++i) {
        if (a[i] == 1) {
          if (a[i + 1] == 1) {
            ubd(i - 1);
          } else if (a[i + 2] == 1) {
            ubd(i);
          } else {
            ubd(i);
            ubd(i - 1);
          }
        }
      }
      for (int i = u; i >= 1; --i) {
        if (a[i] == 1) {
          if (a[i - 1] == 1) {
            ubd(i - 1);
          } else if (a[i - 2] == 1) {
            ubd(i - 2);
          } else {
            ubd(i - 2);
            ubd(i - 1);
          }
        }
      }
    }
    for (int i = 1; i <= n; ++i) {
      assert(a[i] == 0);
    }
    assert(((int)(ans).size()) <= n);
    cout << "YES\n";
    cout << ((int)(ans).size()) << "\n";
    if (!ans.empty()) {
      for (int i = 0; i < ((int)(ans).size()); ++i) cout << ans[i] << ' ';
      cout << "\n";
    }
    return true;
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int tt = 1;
    cin >> tt;
    while (tt--) solv();
    return 0;
  }
  