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 int power(long long int x, long long int y, long long int m) {
    if (y == 0) return 1;
    long long int p = power(x, y / 2, m) % m;
    p = (p * p) % m;
    return (y % 2 == 0) ? p : (x * p) % m;
  }
  vector<long long int> Fac(1, 1), Mod_Inv(1, 1);
  int Last_Index = 0;
  long long int nCr(long long int n, long long int r, long long int m) {
    if (n < 0 || r < 0) return 0;
    if (r > n) return 0;
    if (n > Last_Index) {
      for (long long int i = Last_Index + 1; i <= n; ++i) {
        Fac.push_back(Fac[i - 1] * i);
        Fac[i] %= m;
        Mod_Inv.push_back(power(Fac[i], m - 2, m));
      }
      Last_Index = n;
    }
    return (((Fac[n] * Mod_Inv[r]) % m) * Mod_Inv[n - r]) % m;
  }
  long long int gcd(long long int a, long long int b) {
    return (b ? gcd(b, a % b) : a);
  }
  int Google_Test;
  void solve() {
    long long int n;
    cin >> n;
    vector<long long int> v(n);
    for (long long int i = 0; i < n; ++i) cin >> v[i];
    long long int cnt = 0;
    vector<long long int> cntl(n);
    cnt = 0;
    for (long long int i = 0; i < n; ++i) {
      if (v[i]) ++cnt;
      cntl[i] = cnt;
    }
    vector<long long int> ans;
    {
      queue<long long int> q;
      long long int z = 0;
      for (long long int i = 0; i < n; ++i) {
        if (v[i] == 1)
          q.push(i);
        else
          z++;
      }
      while (!q.empty()) {
        long long int index = q.front();
        q.pop();
        if (index - 2 >= 0 && v[index - 2] == 0 && v[index - 1] == 0 && z > 2 &&
            cntl[index - 2] % 2) {
          ans.push_back(index - 2);
          v[index - 1] = 1;
          v[index - 2] = 1;
          z -= 2;
          q.push(index - 1);
          q.push(index - 2);
        }
        if (index + 2 < n && v[index + 2] == 0 && v[index + 1] == 0 && z > 2 &&
            cntl[index] % 2) {
          ans.push_back(index);
          v[index + 1] = 1;
          v[index + 2] = 1;
          z -= 2;
          q.push(index + 1);
          q.push(index + 2);
        }
      }
    }
    queue<long long int> q;
    for (long long int i = 0; i < n; ++i) {
      if (v[i] == 0) q.push(i);
    }
    while (!q.empty()) {
      long long int index = q.front();
      q.pop();
      if (index - 1 >= 0 && index + 1 < n && v[index - 1] == 1 &&
          v[index + 1] == 1 && cntl[index - 1] % 2) {
        ans.push_back(index - 1);
        v[index - 1] = 0;
        v[index + 1] = 0;
        q.push(index - 1);
        q.push(index + 1);
      }
      if (index - 2 >= 0 && v[index - 2] == 1 && v[index - 1] == 1) {
        ans.push_back(index - 2);
        v[index - 1] = 0;
        v[index - 2] = 0;
        q.push(index - 1);
        q.push(index - 2);
      }
      if (index + 2 < n && v[index + 2] == 1 && v[index + 1] == 1) {
        ans.push_back(index);
        v[index + 1] = 0;
        v[index + 2] = 0;
        q.push(index + 1);
        q.push(index + 2);
      }
    }
    bool f = true;
    for (long long int i : v) {
      if (i == 1) f = false;
    }
    if (f) {
      cout << "YES" << '\n';
      cout << (long long int)ans.size() << '\n';
      if ((long long int)ans.size() != 0) {
        for (long long int i : ans) cout << i + 1 << " ";
        cout << '\n';
      }
    } else {
      cout << "NO" << '\n';
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t;
    cin >> t;
    for (long long int i = 1; i <= t; ++i) {
      solve();
    }
    return 0;
  }
  