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("O2,unroll-loops")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,mmx,avx,avx2")
  using namespace std;
  const long double PI = 3.14159265358979323846;
  const long double E = 2.71828182845904523536;
  const long long LINF = (long long)2e18;
  const int INF = (1 << 30);
  const long double eps = 1e-10;
  const int MAXN = 2e5 + 5;
  int a[MAXN];
  int b[MAXN];
  int l[MAXN];
  int s[MAXN], f[MAXN];
  int inq[MAXN];
  int n;
  vector<int> ans;
  int idx = 1;
  bool remove(int s, int f, vector<int>& skip) {
    if (s == 0 && f == n - 1 && !skip.size()) {
      return false;
    }
    int tct = 0;
    for (int i = s; i <= f; i++) {
      b[i] = 1;
      tct++;
    }
    idx++;
    queue<int> q;
    if (s != 0) {
      q.push(s - 1);
      inq[s - 1] = idx;
    }
    if (f != n - 1) {
      q.push(f - 1);
      inq[f - 1] = idx;
    }
    for (int i : skip) {
      if (inq[i - 1] != idx) {
        q.push(i - 1);
        inq[i - 1] = idx;
      }
      b[i] = 0;
      tct--;
    }
    for (int i = s; i <= f; i++) {
      if (a[i] == 0 && b[i] != 0) {
        a[i + 1] = 1;
        ans.push_back(i);
      }
    }
    while (q.size()) {
      int k = q.front();
      q.pop();
      int ct = 0;
      if (b[k]) {
        ct++;
      }
      if (b[k + 1]) {
        ct++;
      }
      if (b[k + 2]) {
        ct++;
      }
      if (ct != 2) {
        continue;
      }
      tct -= 2;
      b[k] = 0;
      b[k + 1] = 0;
      b[k + 2] = 0;
      ans.push_back(k + 1);
      if (inq[k - 1] != idx && k - 1 >= 0 && k - 1 <= n - 1) {
        inq[k - 1] = idx;
        q.push(k - 1);
      };
      if (inq[k + 1] != idx && k + 1 >= 0 && k + 1 <= n - 1) {
        inq[k + 1] = idx;
        q.push(k + 1);
      };
      if (inq[k - 2] != idx && k - 2 >= 0 && k - 2 <= n - 1) {
        inq[k - 2] = idx;
        q.push(k - 2);
      };
      if (inq[k + 2] != idx && k + 2 >= 0 && k + 2 <= n - 1) {
        inq[k + 2] = idx;
        q.push(k + 2);
      };
    }
    return !tct;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    std::cout << std::fixed << std::setprecision(20);
    int qr;
    for (cin >> qr; qr; qr--) {
      cin >> n;
      ans.clear();
      int ct = 0;
      bool ok = 0;
      for (int i = 0; i < n; i++) {
        cin >> a[i];
        ct ^= a[i];
        ok |= !a[i];
      }
      if (ct || !ok) {
        cout << "NO\n";
        continue;
      }
      a[n] = 0;
      int ptr = 0;
      for (int i = 0; i < n;) {
        int j = i;
        l[ptr] = 0;
        s[ptr] = i;
        while (j < n && a[j] == a[i]) {
          l[ptr]++;
          j++;
        }
        f[ptr] = j - 1;
        i = j;
        ptr++;
      }
      for (int i = 0; i < ptr;) {
        if (!a[s[i]]) {
          i++;
          continue;
        }
        int temp = l[i];
        vector<int> skip;
        int j = i;
        while (temp & 1) {
          int k0 = j + 1;
          temp += l[k0] - (l[k0] & 1);
          if (l[k0] & 1) {
            skip.push_back(f[k0]);
          }
          int k1 = j + 2;
          if (k1 == ptr || k0 == ptr) {
            cout << "NO\n";
            goto done;
          }
          temp += l[k1];
          j += 2;
        }
        if (!remove(s[i], f[j], skip)) {
          cout << "NO\n";
          goto done;
        }
        i = j + 1;
      }
      cout << "YES\n";
      cout << ans.size() << '\n';
      for (int i : ans) {
        cout << i << ' ';
      }
      cout << '\n';
    done:;
    }
  }
  